leetcode笔记

Longest Palindromic Substring

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Given a string s, find the longest palindromic substring in s. You may assume that the maximum length of s is 1000.

Example 1:

Input: “babad”
Output: “bab”
Note: “aba” is also a valid answer.
Example 2:

Input: “cbbd”
Output: “bb”

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思路（动态规划）：

• 创建二维数组dp[i][j]，行列值为字符串的size()，其中i，j分别代表字符串的起始和终止索引。dp[i][j] = 1代表字符串s[i][j]为回文。
• 分为三种情况:
  - dp[i][i] = 1，即单个字符为回文
  - dp[i][i+1]=1，两个字符相同为回文
  - dp[i][j+len]=1，len = 3,4,5…
• 遍历每个len的基础上，遍历 i - > j 的字符，并判断否s[i]== s[j] && dp[i+1][j-1] == 1

代码实现：

class Solution {
public:
    string longestPalindrome(string s) {
        int length = s.size();
        int max = 0;
        int start =0;
        vector< vector<int>> dp(length,vector<int>(length));

        for (int i=0; i<length; i++){
            dp[i][i] = 1;
            if (s[i]==s[i+1]){
                dp[i][i+1]=1;

                max = 2;
                start = i;
            }
        }

        for (int len=3;len<length;len++){
            for(int i=0;i+len<length;i++){
                int j = i+len;

                if(s[i]==s[j] && dp[i+1][j-1]){
                    dp[i][j]=1;
                    max = len;
                    start = i;
                }
            }
        }
        return s.substr(start, start+max);
    }
};

Valid Parentheses（有效的括号）

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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

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思路（利用栈）：

• 遍历字符串，遇到左括号压入栈，遇到右括号的话取top()出来比较(通过ASCII比较)，比较符合的话pop出栈。
• 特殊情况：字符串为奇数；栈为空就遇到右括号。

#include<stack>
#include<string>
class Solution {
public:
	bool isValid(string s) {
		stack<int> st;
		if (s.length() % 2 != 0){
			return false;
		}
		for (int i = 0; i < s.length(); i++){
			if (s[i] == '}' || s[i] == ']' || s[i] == ')' ){
				if (st.empty()) return false;
				else if(s[i]-st.top()==1 || s[i]-st.top()==2){  // ASCII {}相差2
                    st.pop();
                }
			}
			else st.push(s[i]);
		}
		return st.empty();
	}
};