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本系列为笔者的 Leetcode 刷题记录，顺序为 Hot 100 题官方顺序，根据标签命名，记录笔者总结的做题思路，附部分代码解释和疑问解答，01~07为C++语言，08及以后为Java语言。

01 有效的括号

class Solution {public boolean isValid(String s) {int n = s.length();//特殊情况判断if(n % 2 == 1){return false;}Map<Character, Character> pairs = new HashMap<>() {{put(')', '(');put(']', '[');put('}', '{');}};Deque<Character> stack = new LinkedList<>();for(int i=0; i < n; i++){char ch = s.charAt(i);if(pairs.containsKey(ch)){ //右括号，做判断if(stack.isEmpty() || stack.peek() != pairs.get(ch)){return false;}stack.pop();}else{ //左括号，压入栈stack.push(ch);}}return stack.isEmpty();}
}

02 最小栈

class MinStack {public MinStack() {}public void push(int val) {}public void pop() {}public int top() {}public int getMin() {}
}/*** Your MinStack object will be instantiated and called as such:* MinStack obj = new MinStack();* obj.push(val);* obj.pop();* int param_3 = obj.top();* int param_4 = obj.getMin();*/

方法：辅助栈

class MinStack {Deque<Integer> xStack;Deque<Integer> minStack;public MinStack() {xStack = new LinkedList<>();minStack = new LinkedList<>();minStack.push(Integer.MAX_VALUE);}public void push(int val) {xStack.push(val);minStack.push(Math.min(minStack.peek(), val));}public void pop() {xStack.pop();minStack.pop();}public int top() {return xStack.peek();}public int getMin() {return minStack.peek();}
}/*** Your MinStack object will be instantiated and called as such:* MinStack obj = new MinStack();* obj.push(val);* obj.pop();* int param_3 = obj.top();* int param_4 = obj.getMin();*/

03 字符串解码

class Solution {int ptr;public String decodeString(String s) {LinkedList<String> stk = new LinkedList<>();ptr = 0;while(ptr < s.length()){char cur = s.charAt(ptr);if(Character.isDigit(cur)){String digits = getDigit(s);stk.addLast(digits); //1. 添加数字}else if(Character.isLetter(cur) || cur == '['){stk.addLast(String.valueOf(s.charAt(ptr++))); //2.添加单个字符}else{//获取括号中的字符串ptr++;LinkedList<String> sub = new LinkedList<>();while(!"[".equals(stk.peekLast())){sub.addLast(stk.removeLast());}Collections.reverse(sub);stk.removeLast(); //弹出左括号int time = Integer.parseInt(stk.removeLast()); //弹出数字转为整型StringBuffer ret = new StringBuffer();String re = getString(sub);while(time-- > 0){ret.append(re);}stk.addLast(ret.toString()); //3.添加重复字符串}}return getString(stk);}public String getDigit(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();}
}

04 每日温度

class Solution {public int[] dailyTemperatures(int[] temperatures) {int length = temperatures.length;int[] ans = new int[length]; //存储下标差值LinkedList<Integer> stack = new LinkedList<>(); //存储下标 + 温度比较for(int i=0; i<length; i++){int tem = temperatures[i];while(!stack.isEmpty() && tem > temperatures[stack.peek()]){int preIndex = stack.pop(); ans[preIndex] = i - preIndex; //⭐}stack.push(i);}return ans;}
}

05 柱状图的最大矩形

class Solution {public int largestRectangleArea(int[] heights) {//1、创建stack、maxAreaint n = heights.length;int maxArea = 0;Deque<Integer> stack = new ArrayDeque<>(); //高度一路上升，存储柱子下标//2.for循环、遍历柱子下标for(int i=0; i<=n; i++){int currHeight = (i == n) ? 0 : heights[i];while(!stack.isEmpty() && currHeight < heights[stack.peek()]){int height = heights[stack.pop()]; //⭐int width = stack.isEmpty() ? i : i - stack.peek() - 1;maxArea = Math.max(maxArea, height * width);}stack.push(i);}//3.返回最大面积return maxArea;}
}