【Leetcode】算法总结——栈
1. 栈
- Valid Parentheses（有效的括号）
Given a string containing just the characters ‘(’, ‘)’, ‘{’, ‘}’, ‘[’ and ‘]’, determine if the input string is valid.
An input string is valid if:
Open brackets must be closed by the same type of brackets.
Open brackets must be closed in the correct order.
Note that an empty string is also considered valid.
Example 1:
Input: “()”
Output: true
Example 2:
Input: “()[]{}”
Output: true
Example 3:
Input: “(]”
Output: false
Example 4:
Input: “([)]”
Output: false
Example 5:
Input: “{[]}”
Output: true
```java
class Solution {
public boolean isValid(String s) {
Stack<Character> stack=new Stack<Character>();
int len=s.length();
for(int i=0;i<len;i++) {
if(s.charAt(i)=='['||s.charAt(i)=='('||s.charAt(i)=='{') {
stack.push(s.charAt(i));
}else if (s.charAt(i)==']') {
//从栈中取出一个元素
if(stack.empty()==true){
return false;
}
char ch=stack.pop();
if(ch!='[') {
return false;
}
}else if(s.charAt(i)==')') {
if(stack.empty()==true){
return false;
}
char ch=stack.pop();
if(ch!='(') {
return false;
}
}else if(s.charAt(i)=='}'){
if(stack.empty()==true){
return false;
}
char ch=stack.pop();
if(ch!='{') {
return false;
}
}
}
if(stack.empty()==true) {
return true;
}
return false;
}
}
- Trapping Rain Water（接雨水）
Given n non-negative integers representing an elevation map where the width of each bar is 1, compute how much water it is able to trap after raining.
Example:
Input: [0,1,0,2,1,0,1,3,2,1,2,1]
Output: 6
思路：
使用一个栈，栈中元素是数组height的序号
遍历数组
当当前元素大于栈元素时，出栈
计算两堵墙height【i】与height【栈顶元素】的最小值min
计算两堵墙的之间的距离distance。
面积为 distance*min
class Solution {
public int trap(int[] height) {
Stack<Integer> stack=new Stack<Integer>();
int sum=0;
int len=height.length;
for(int i=0;i<len;i++){
while(!stack.empty()&&height[stack.peek()]<height[i]) {
//出栈
int no=stack.pop();
if(stack.empty()) {
break;
}
int distance=i-stack.peek()-1;
int min=height[i]<height[stack.peek()]?height[i]:height[stack.peek()];
sum+=distance*(min-height[no]);
}
//将当前的入栈
stack.push(i);
}
return sum;
}
}
