力扣：面试题30:包含min函数的栈
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//定义栈的数据结构，请在该类型中实现一个能够得到栈的最小元素的 min 函数在该栈中，调用 min、push 及 pop 的时间复杂度都是 O(1)。
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//
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//
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//
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// 示例:
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//
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// MinStack minStack = new MinStack();
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//minStack.push(-2);
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//minStack.push(0);
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//minStack.push(-3);
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//minStack.min(); --> 返回 -3.
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//minStack.pop();
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//minStack.top(); --> 返回 0.
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//minStack.min(); --> 返回 -2.
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//
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//
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//
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//
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// 提示：
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//
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//
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// 各函数的调用总次数不超过 20000 次
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//
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//
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//
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//
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// 注意：本题与主站 155 题相同：https://leetcode-cn.com/problems/min-stack/
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// Related Topics 栈 设计
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// 👍 114 👎 0
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package leetcode.editor.cn;
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import java.util.Stack;
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//面试题30:包含min函数的栈
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public class BaoHanMinhanShuDeZhanLcof{
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public static void main(String[] args) {
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//测试代码
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// Solution solution = new BaoHanMinhanShuDeZhanLcof().new Solution();
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}
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//力扣代码
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//leetcode submit region begin(Prohibit modification and deletion)
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class MinStack {
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Stack<Integer> stack;
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Stack<Integer> min;
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public MinStack() {
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stack = new Stack<>();
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min = new Stack<>();
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}
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public void push(int val) {
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if (stack.isEmpty()) {
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min.push(val);
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} else {
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min.push(Math.min(min.peek(), val));
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}
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stack.push(val);
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}
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public void pop() {
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stack.pop();
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min.pop();
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}
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public int top() {
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return stack.peek();
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}
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public int getMin() {
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return min.peek();
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}
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}
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/**
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* Your MinStack object will be instantiated and called as such:
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* MinStack obj = new MinStack();
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* obj.push(x);
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* obj.pop();
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* int param_3 = obj.top();
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* int param_4 = obj.min();
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*/
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//leetcode submit region end(Prohibit modification and deletion)
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}
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@ -0,0 +1,28 @@
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<p>定义栈的数据结构，请在该类型中实现一个能够得到栈的最小元素的 min 函数在该栈中，调用 min、push 及 pop 的时间复杂度都是 O(1)。</p>
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<p> </p>
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<p><strong>示例:</strong></p>
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<pre>MinStack minStack = new MinStack();
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minStack.push(-2);
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minStack.push(0);
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minStack.push(-3);
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minStack.min(); --> 返回 -3.
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minStack.pop();
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minStack.top(); --> 返回 0.
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minStack.min(); --> 返回 -2.
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</pre>
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<p> </p>
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<p><strong>提示：</strong></p>
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<ol>
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<li>各函数的调用总次数不超过 20000 次</li>
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</ol>
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<p> </p>
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<p>注意：本题与主站 155 题相同：<a href="https://leetcode-cn.com/problems/min-stack/">https://leetcode-cn.com/problems/min-stack/</a></p>
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<div><div>Related Topics</div><div><li>栈</li><li>设计</li></div></div>\n<div><li>👍 114</li><li>👎 0</li></div>
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@ -68,7 +68,7 @@ public class FuZaLianBiaoDeFuZhiLcof {
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List<Integer> var = Arrays.asList(7, 13, 11, 10, 1);
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List<Integer> random = Arrays.asList(null, 0, 4, 2, 0);
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Node head = new Node(0);
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head = head.setHead(var, random);
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// head = head.setHead(var, random);
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solution.copyRandomList(head);
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}
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@ -47,13 +47,13 @@ import java.util.Stack;
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public class MinStack {
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public static void main(String[] args) {
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//测试代码
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// Solution solution = new MinStack().new Solution();
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// Solution solution = new MinStack1().new Solution();
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}
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//力扣代码
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//leetcode submit region begin(Prohibit modification and deletion)
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class MinStack {
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class MinStack1 {
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// Stack<Integer> stack;
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// Stack<Integer> min;
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// public MinStack() {
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@ -101,7 +101,7 @@ public class MinStack {
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}
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Stack<Data> stack;
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public MinStack() {
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public MinStack1() {
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stack = new Stack<>();
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}
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---
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### 🧠 解题思路
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我们解决这道题的关键在于，需要知道哪些是需要去除的外层括号，为了找到这些需要去除的外层括号，我们可以使用到计数器。
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**规则：** 遇到左括号，我们的计数器 *+1*，遇到右括号，我们的计数器 *-1*。
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这样的话，一组连续且有效的括号，将不会对计数器的值产生变化。
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```js
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// 示例一
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当前的计数值: 0 1 0 1
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( ) ( )
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遍历后计数值: 1 0 1 0
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// 示例二
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当前的计数值: 0 1 2 1 2 1 0 1
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( ( ) ( ) ) ( )
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遍历后计数值: 1 2 1 2 1 0 1 0
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```
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根据上述两个示例，我们可以很快的找出规律：
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1. 遇到左括号，当前计数值大于 *0* ，则属于有效的左括号。
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2. 遇到右括号，当前计数值大于 *1* ，则属于有效的右括号。
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---
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### 🎨 图解演示
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---
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### 🍭 示例代码
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```Javascript []
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var removeOuterParentheses = function(S) {
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let count = 0, ans = '';
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for (let i = 0; i < S.length; i++) {
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if(S[i] === '(' && count++ > 0) ans += '('
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if(S[i] === ')' && count-- > 1) ans += ')';
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}
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return ans;
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};
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```
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---
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### 转身挥手
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嘿，少年，做图不易，留下个赞或评论再走吧！谢啦~ 💐
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差点忘了，祝你牛年大吉 🐮 ，AC 和 Offer 📑 多多益善~
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⛲⛲⛲ 期待下次再见~
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@ -0,0 +1,48 @@
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---
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### 🧠 解题思路
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分析题意之后，可以得出以下结论：
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1. 字符要做比较，所以之前的字符应该被存储下来，这里我们会用到栈。
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2. 遍历字符，若栈顶和当前字符正好大小写都具备，则弹出栈顶抵消，否则当前字符入栈。
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---
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### 🎨 图解演示
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---
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### 🍭 示例代码
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```Javascript []
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var makeGood = function(s) {
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let res = [];
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for(let i of s){
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if(
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res.length &&
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res[res.length - 1] !== i &&
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res[res.length - 1].toUpperCase() === i.toUpperCase()
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){
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res.pop();
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}else{
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res.push(i);
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}
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}
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return res.join("");
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};
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```
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---
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### 转身挥手
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嘿，少年，做图不易，留下个赞或评论再走吧！谢啦~ 💐
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差点忘了，祝你牛年大吉 🐮 ，AC 和 Offer 📑 多多益善~
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⛲⛲⛲ 期待下次再见~
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