比特微光

栈的解题模版

/////////////////////////////
// push(): 将元素添加到栈的头部
// pop(): 从栈的头部移除元素
// peek(): 获取栈顶元素

// addFirst();
// removeFirst();

// addLast();
// removeLast();

// pollFirst();
// offerFirst();
// peekFirst();

// pollLast();
// offerLast();
// peekLast();
////////////////////////////

// 1. StringBuffer
StringBuffer stack = new StringBuffer();
int top = -1;
stack.deleteCharAt(top);
stack.stack.append(ch);

// 2. LinkedList
Deque<Integer> stack = new LinkedList<Integer>();
stack.push(x);
stack.pop();
stack.peek();

// 3. ArrayDeque
Deque<Integer> stack = new ArrayDeque<Integer>();
stack.push(x);
stack.pop();
stack.peek();

// 4. Stack
Stack<Character> stack = new Stack<>();
stack.push(x);
stack.pop();
stack.peek();

// 5. 
List<String> array = new ArrayList();
Collections.reverse(sub);

删除字符串中所有重复项

class Solution {
    public String removeDuplicates(String s) {
        StringBuffer stack = new StringBuffer();
        int top = -1;
      	// 1. 遍历字符串字符
        for(int i = 0; i < s.length(); i++) {
            char ch = s.charAt(i);
          	// 2. 栈顶元素 等于 遍历字符，则将栈顶元素出栈
            if (top >= 0 && stack.charAt(top) == ch) {
                stack.deleteCharAt(top);
                --top;
            } else {
              	// 3. 否则，遍历字符入栈
                stack.append(ch);
                ++top;
            }
        }
      	// 将栈内元素输出
        return stack.toString();
    }
}

最小栈

// 只需要设计一个数据结构，使得每个元素 a 与其相应的最小值 m 时刻保持一一对应。因此我们可以使用一个辅助栈，与元素栈同步插入与删除，用于存储与每个元素对应的最小值。

class MinStack {
    Deque<Integer> xStack;
    Deque<Integer> minStack;

    public MinStack() {
        xStack = new LinkedList<Integer>();
        minStack = new LinkedList<Integer>();
      	// 细节：将栈底元素初始化为最大值
        minStack.push(Integer.MAX_VALUE);
    }
    
    public void push(int x) {
        xStack.push(x);
        minStack.push(Math.min(minStack.peek(), x));
    }
    
    public void pop() {
        xStack.pop();
        minStack.pop();
    }
    
    public int top() {
        return xStack.peek();
    }
    
    public int getMin() {
        return minStack.peek();
    }
}

有效的括号

class Solution {
    public boolean isValid(String s) {
        int n = s.length();
        if (n % 2 == 1) {
            // 1. 如果字符串长度是奇数，则一定不可能是有效括号
            return false;
        }

        // 2. 使用一个map字典，用来保存左右括号之间的映射关系
        Map<Character, Character> pairs = new HashMap<Character, Character>() {{
            put(')', '(');
            put(']', '[');
            put('}', '{');
        }};
        Deque<Character> stack = new LinkedList<Character>();
        for (int i = 0; i < n; i++) {
            char ch = s.charAt(i);
            if (pairs.containsKey(ch)) {
                // 3. 如果当前字符是右括号，则：如果栈顶元素不是坐括号、或者栈为空，则一定不是有效括号
                if (stack.isEmpty() || stack.peek() != pairs.get(ch)) {
                    return false;
                }
                stack.pop();
            } else {
                stack.push(ch);
            }
        }
        return stack.isEmpty();
    }
}

基本计算器 II

class Solution {
    public int calculate(String s) {
        Deque<Integer> stack = new ArrayDeque<Integer>();
        char preSign = '+';
        int num = 0;
        int n = s.length();
        for (int i = 0; i < n; ++i) {
            if (Character.isDigit(s.charAt(i))) {
                num = num * 10 + s.charAt(i) - '0';
            }
            if (!Character.isDigit(s.charAt(i)) && s.charAt(i) != ' ' || i == n - 1) {
                switch (preSign) {
                case '+':
                    stack.push(num);
                    break;
                case '-':
                    stack.push(-num);
                    break;
                case '*':
                    stack.push(stack.pop() * num);
                    break;
                default:
                    stack.push(stack.pop() / num);
                }
                preSign = s.charAt(i);
                num = 0;
            }
        }
        int ans = 0;
        while (!stack.isEmpty()) {
            ans += stack.pop();
        }
        return ans;
    }
}

每日温度

class Solution {
    public int[] dailyTemperatures(int[] temperatures) {
        int length = temperatures.length;
        int[] ans = new int[length];
        Deque<Integer> stack = new LinkedList<Integer>();
        for (int i = 0; i < length; i++) {
            int temperature = temperatures[i];
            while (!stack.isEmpty() && temperature > temperatures[stack.peek()]) {
                int prevIndex = stack.pop();
                ans[prevIndex] = i - prevIndex;
            }
            stack.push(i);
        }
        return ans;
    }
}

逆波兰表达式求值

class Solution {
    public int evalRPN(String[] tokens) {
        Deque<Integer> stack = new LinkedList<Integer>();
        int n = tokens.length;
        for (int i = 0; i < n; i++) {
            String token = tokens[i];
            if (isNumber(token)) {
                stack.push(Integer.parseInt(token));
            } else {
                int num2 = stack.pop();
                int num1 = stack.pop();
                switch (token) {
                    case "+":
                        stack.push(num1 + num2);
                        break;
                    case "-":
                        stack.push(num1 - num2);
                        break;
                    case "*":
                        stack.push(num1 * num2);
                        break;
                    case "/":
                        stack.push(num1 / num2);
                        break;
                    default:
                }
            }
        }
        return stack.pop();
    }

    public boolean isNumber(String token) {
        return !("+".equals(token) || "-".equals(token) || "*".equals(token) || "/".equals(token));
    }
}

用栈实现队列

class MyQueue {
    Deque<Integer> inStack;
    Deque<Integer> outStack;

    public MyQueue() {
        inStack = new ArrayDeque<Integer>();
        outStack = new ArrayDeque<Integer>();
    }

    public void push(int x) {
        inStack.push(x);
    }

    public int pop() {
        if (outStack.isEmpty()) {
            in2out();
        }
        return outStack.pop();
    }

    public int peek() {
        if (outStack.isEmpty()) {
            in2out();
        }
        return outStack.peek();
    }

    public boolean empty() {
        return inStack.isEmpty() && outStack.isEmpty();
    }

    private void in2out() {
        while (!inStack.isEmpty()) {
            outStack.push(inStack.pop());
        }
    }
}

字符串解码

class Solution {
    int ptr;

    public String decodeString(String s) {
        LinkedList<String> stk = new LinkedList<String>();
        ptr = 0;

        while (ptr < s.length()) {
            char cur = s.charAt(ptr);
            if (Character.isDigit(cur)) {
                // 获取一个数字并进栈
                String digits = getDigits(s);
                stk.addLast(digits);
            } else if (Character.isLetter(cur) || cur == '[') {
                // 获取一个字母并进栈
                stk.addLast(String.valueOf(s.charAt(ptr++))); 
            } else {
                ++ptr;
                LinkedList<String> sub = new LinkedList<String>();
                while (!"[".equals(stk.peekLast())) {
                    sub.addLast(stk.removeLast());
                }
                Collections.reverse(sub);
                // 左括号出栈
                stk.removeLast();
                // 此时栈顶为当前 sub 对应的字符串应该出现的次数
                int repTime = Integer.parseInt(stk.removeLast());
                StringBuffer t = new StringBuffer();
                String o = getString(sub);
                // 构造字符串
                while (repTime-- > 0) {
                    t.append(o);
                }
                // 将构造好的字符串入栈
                stk.addLast(t.toString());
            }
        }

        return getString(stk);
    }

    public String getDigits(String s) {
        StringBuffer ret = new StringBuffer();
        while (Character.isDigit(s.charAt(ptr))) {
            ret.append(s.charAt(ptr++));
        }
        return ret.toString();
    }

    public String getString(LinkedList<String> v) {
        StringBuffer ret = new StringBuffer();
        for (String s : v) {
            ret.append(s);
        }
        return ret.toString();
    }
}

最长有效括号

class Solution {
    public int longestValidParentheses(String s) {
        int maxans = 0;
        Deque<Integer> stack = new LinkedList<Integer>();
        stack.push(-1);
        for (int i = 0; i < s.length(); i++) {
            if (s.charAt(i) == '(') {
                stack.push(i);
            } else {
                stack.pop();
                if (stack.isEmpty()) {
                    stack.push(i);
                } else {
                    maxans = Math.max(maxans, i - stack.peek());
                }
            }
        }
        return maxans;
    }
}

验证栈序列

class Solution {
    public boolean validateStackSequences(int[] pushed, int[] popped) {
        Deque<Integer> stack = new ArrayDeque<Integer>();
        int n = pushed.length;
        for (int i = 0, j = 0; i < n; i++) {
            stack.push(pushed[i]);
            while (!stack.isEmpty() && stack.peek() == popped[j]) {
                stack.pop();
                j++;
            }
        }
        return stack.isEmpty();
    }
}

从尾到头打印链表

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public int[] reverseBookList(ListNode head) {
        Deque<Integer> stack = new LinkedList();

        while (head != null) {
            stack.push(head.val);
            head = head.next;
        }
        int[] res = new int[stack.size()];
        for (int i = 0; i < res.length; i++) {
            res[i] = stack.pop();
        }
        return res;

    }
}

简化路径

/////////////栈的相关操作////////////
class Solution {
    public String simplifyPath(String path) {
        String[] names = path.split("/");
        Deque<String> stack = new LinkedList();

        // 1. 对入参使用“/”分割后的数组遍历；
        for (String name : names) {
            if ("..".equals(name)) {
                if (!stack.isEmpty()) {
                    // 1.2 如果等于“..” 且 栈非空，则出栈
                    stack.pop();
                }
            // 1.1 如果 非空 且 不等于 “.” , 则入栈  
            } else if (name.length() > 0 && !".".equals(name)) {
                stack.push(name);
            }
        }
        StringBuffer ans = new StringBuffer();
        if (stack.isEmpty()) {
            ans.append('/');
        } else {
            while (!stack.isEmpty()) {
                ans.append('/');
              	// 注意：从栈底开始遍历元素
                ans.append(stack.removeLast());
            }
        }
        return ans.toString();
    }
}

接雨水

// 单调栈解法
class Solution {
    public int trap(int[] height) {
        int ans = 0;
      	// 单调栈存储的是下标，满足从栈底到栈顶的下标对应的数组 height 中的元素递减。
        Deque<Integer> stack = new LinkedList<Integer>();
        int n = height.length;
        for (int i = 0; i < n; ++i) {
          	// 栈非空 且 数组当前元素 大于 栈顶元素，则出栈
            while (!stack.isEmpty() && height[i] > height[stack.peek()]) {
              	// 1. 栈中只有一个元素
                int top = stack.pop();
                if (stack.isEmpty()) {
                    break;
                }
              	// 2. 
                int left = stack.peek();
                int currWidth = i - left - 1;
                int currHeight = Math.min(height[left], height[i]) - height[top];
                ans += currWidth * currHeight;
            }
          	// 否则，则入栈；
            stack.push(i);
        }
        return ans;
    }
}

//

最大矩阵

class Solution {
    public int maximalRectangle(char[][] matrix) {
        int m = matrix.length;
        if (m == 0) {
            return 0;
        }
        int n = matrix[0].length;
        int[][] left = new int[m][n];

        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (matrix[i][j] == '1') {
                    left[i][j] = (j == 0 ? 0 : left[i][j - 1]) + 1;
                }
            }
        }

        int ret = 0;
        for (int j = 0; j < n; j++) { // 对于每一列，使用基于柱状图的方法
            int[] up = new int[m];
            int[] down = new int[m];

            Deque<Integer> stack = new LinkedList<Integer>();
            for (int i = 0; i < m; i++) {
                while (!stack.isEmpty() && left[stack.peek()][j] >= left[i][j]) {
                    stack.pop();
                }
                up[i] = stack.isEmpty() ? -1 : stack.peek();
                stack.push(i);
            }
            stack.clear();
            for (int i = m - 1; i >= 0; i--) {
                while (!stack.isEmpty() && left[stack.peek()][j] >= left[i][j]) {
                    stack.pop();
                }
                down[i] = stack.isEmpty() ? m : stack.peek();
                stack.push(i);
            }

            for (int i = 0; i < m; i++) {
                int height = down[i] - up[i] - 1;
                int area = height * left[i][j];
                ret = Math.max(ret, area);
            }
        }
        return ret;
    }
}

和至少为k的最短子数组

class Solution {
    public int shortestSubarray(int[] nums, int k) {
        int n = nums.length;
        long[] preSumArr = new long[n + 1];
        for (int i = 0; i < n; i++) {
            preSumArr[i + 1] = preSumArr[i] + nums[i];
        }
        int res = n + 1;
        Deque<Integer> queue = new ArrayDeque<Integer>();
        for (int i = 0; i <= n; i++) {
            long curSum = preSumArr[i];
            while (!queue.isEmpty() && curSum - preSumArr[queue.peekFirst()] >= k) {
                res = Math.min(res, i - queue.pollFirst());
            }
            while (!queue.isEmpty() && preSumArr[queue.peekLast()] >= curSum) {
                queue.pollLast();
            }
            queue.offerLast(i);
        }
        return res < n + 1 ? res : -1;
    }
}

去除重复字母

class Solution {
    public String removeDuplicateLetters(String s) {
        boolean[] vis = new boolean[26];
        int[] num = new int[26];
        for (int i = 0; i < s.length(); i++) {
            num[s.charAt(i) - 'a']++;
        }

        StringBuffer sb = new StringBuffer();
        for (int i = 0; i < s.length(); i++) {
            char ch = s.charAt(i);
            if (!vis[ch - 'a']) {
                while (sb.length() > 0 && sb.charAt(sb.length() - 1) > ch) {
                    if (num[sb.charAt(sb.length() - 1) - 'a'] > 0) {
                        vis[sb.charAt(sb.length() - 1) - 'a'] = false;
                        sb.deleteCharAt(sb.length() - 1);
                    } else {
                        break;
                    }
                }
                vis[ch - 'a'] = true;
                sb.append(ch);
            }
            num[ch - 'a'] -= 1;
        }
        return sb.toString();
    }
}

下一个更大元素 I

// 单调栈+哈希表
class Solution {
    public int[] nextGreaterElement(int[] nums1, int[] nums2) {
        Map<Integer, Integer> map = new HashMap<Integer, Integer>();
        Deque<Integer> stack = new ArrayDeque<Integer>();
        for (int i = nums2.length - 1; i >= 0; --i) {
            int num = nums2[i];
            while (!stack.isEmpty() && num >= stack.peek()) {
                stack.pop();
            }
            map.put(num, stack.isEmpty() ? -1 : stack.peek());
            stack.push(num);
        }
        int[] res = new int[nums1.length];
        for (int i = 0; i < nums1.length; ++i) {
            res[i] = map.get(nums1[i]);
        }
        return res;
    }
}

下一个更大元素 II

class Solution {
    public int[] nextGreaterElements(int[] nums) {
        int n = nums.length;
        int[] ret = new int[n];
        Arrays.fill(ret, -1);
        Deque<Integer> stack = new LinkedList<Integer>();
        for (int i = 0; i < n * 2 - 1; i++) {
            while (!stack.isEmpty() && nums[stack.peek()] < nums[i % n]) {
                ret[stack.pop()] = nums[i % n];
            }
            stack.push(i % n);
        }
        return ret;
    }
}