天池在线编程条件串（DP）

文章目录

- 1. 题目
- 2. 解题

1. 题目

来源：https://tianchi.aliyun.com/oj/164427478262600292/204998627646706401

2. 解题

把字符串分成ace，bdf 两部分进行处理，求以某个字符结束时的最小删除次数

class Solution {
public:/*** @param s: write your code here* @return: The longest length that satisfies the condition*/int conditionString(string &s) {// write your code herestring s1, s2;for(char c : s){if(c == 'a' || c == 'c' || c == 'e')s1 += c;elses2 += char(c-1);}return s.size()-f(s1)-f(s2);}int f(string &s){int n = s.size();if(n == 0)return 0;vector<vector<int>> dp(n,vector<int>(3, INT_MAX));// 求以某个字符结尾时,满足要求的,删除的最小次数dp[0][0] = s[0]=='a' ? 0 : 1;dp[0][1] = s[0]=='c' ? 0 : 1;dp[0][2] = s[0]=='e' ? 0 : 1;for(int i = 1; i < n; ++i){if(s[i] == 'a'){dp[i][0] = dp[i-1][0];dp[i][1] = 1+min(dp[i-1][0], dp[i-1][1]);dp[i][2] = 1+min(dp[i-1][0], min(dp[i-1][1], dp[i-1][2]));}else if(s[i] == 'c'){dp[i][0] = 1+dp[i-1][0];dp[i][1] = min(dp[i-1][0], dp[i-1][1]);dp[i][2] = 1+min(dp[i-1][0], min(dp[i-1][1], dp[i-1][2]));}else{dp[i][0] = 1+dp[i-1][0];dp[i][1] = 1+min(dp[i-1][0], dp[i-1][1]);dp[i][2] = min(dp[i-1][0], min(dp[i-1][1], dp[i-1][2]));}}return min(dp[n-1][0],min(dp[n-1][1], dp[n-1][2]));}
};