交錯字串(動態規劃)- leetcode 97
阿新 • • 發佈:2018-12-24
給定三個字串 s1, s2, s3, 驗證 s3 是否是由 s1 和 s2 交錯組成的。
示例 1:
輸入: s1 = "aabcc", s2 = "dbbca", s3 = "aadbbcbcac" 輸出: true
示例 2:
輸入: s1 = "aabcc", s2 = "dbbca", s3 = "aadbbbaccc" 輸出: false
向右走就是s2的這位去匹配s3的當前位,向下走就是s1的這位匹配s3的當前位。
原始碼:GCC
#include <stdio.h> #include <stdlib.h> #include <stdbool.h> #include <string.h> /* * Interleaving string s3 is whether a line from left top to right buttom. * or not. * * s2 ..... *s1 T F F F F F *. T F F F F F *. T T T T T F *. F T T F T F * F F T T T T * F F F T F T */ static bool isInterleave(char* s1, char* s2, char* s3) { int i, j; int len1 = strlen(s1); int len2 = strlen(s2); int len3 = strlen(s3); if (len1 + len2 != len3) { return false; } bool *table = malloc((len1 + 1) * (len2 + 1) * sizeof(bool)); bool **dp = malloc((len1 + 1) * sizeof(bool *)); for (i = 0; i < len1 + 1; i++) { dp[i] = &table[i * (len2 + 1)]; } dp[0][0] = true; for (i = 1; i < len1 + 1; i++) { dp[i][0] = dp[i - 1][0] && s1[i - 1] == s3[i - 1]; } for (i = 1; i < len2 + 1; i++) { dp[0][i] = dp[0][i - 1] && s2[i - 1] == s3[i - 1]; } for (i = 1; i < len1 + 1; i++) { for (j = 1; j < len2 + 1; j++) { bool up = dp[i - 1][j] && s1[i - 1] == s3[i + j - 1]; bool left = dp[i][j - 1] && s2[j - 1] == s3[i + j - 1]; dp[i][j] = up || left; } } return dp[len1][len2]; } int main(int argc, char **argv) { if (argc != 4) { fprintf(stderr, "Usage: ./test s1 s2 s3\n"); exit(-1); } printf("%s\n", isInterleave(argv[1], argv[2], argv[3]) ? "true" : "false"); return 0; }