1015. Letter-moving Game (35)解题报告

原题链接：

1015. Letter-moving Game (35)

解题思路：

经观察可知，S中子序列与T中连续子序列相同的最大长度为L，原序列长度减去L就是最少移动次数。

通过画面：

代码：

#define _CRT_SECURE_NO_WARNINGS#include <cstdio>#include <cstdlib>#include <cassert>#include <string>#include <vector>#include <algorithm>#include <queue>#include <set>#include <iostream>using namespace std;int counting(string S, string T);int main(){string S, T;cin >> S >> T;printf("%d\n", counting(S, T));return 0;}int counting(string S, string T){int steps = 0;int maxSubStr = 1;for (int i = 2; i <= (int)S.length(); i++) {bool flag = true;for (int j = 0; j <= (int)S.length() - i && flag; j++) {for (int k1 = 0, k2 = 0; k1 < (int)S.length() && k2 < i && flag; k1++) {if (S[k1] == T[j + k2]) {k2++;}if (k2 == i) {flag = false;}}}if (!flag) {maxSubStr = i;}else {break;}}steps = S.length() - maxSubStr;return steps;}