1. 题目

来源：https://tianchi.aliyun.com/oj/164423301311799378/184808348725744276

2. 解题

class Solution {
public:/*** @param s: a string* @param t: a string* @param n: max times to swap a 'l' and a 'r'.* @return: return if s can transform to t.*/bool LRString(string &s, string &t, int n) {// write your code here.if(s.size() != t.size())return false;//长度不等不行int ls = 0, rs = 0, lt = 0, rt = 0;vector<int> sLidx, tLidx;for(int i = 0; i < s.size(); i++){if(s[i] == 'l'){ls++;sLidx.push_back(i);}elsers++;}for(int i = 0; i < t.size(); i++){if(t[i] == 'l'){lt++;tLidx.push_back(i);}elsert++;}if(ls != lt || rs != rt)return false;// lr字符数量不等不行int sw = 0;for(int i = 0; i < sLidx.size(); i++){if(sLidx[i] == tLidx[i])continue;else if(sLidx[i] < tLidx[i])return false;// l 的位置只能往左，不可能做到elsesw += sLidx[i] - tLidx[i];// 要移动的距离}return sw <= n;}
};

50ms C++

---

我的CSDN博客地址 https://michael.blog.csdn.net/

长按或扫码关注我的公众号（Michael阿明），一起加油、一起学习进步！