题目: 编辑距离*
目前感受下来,这种字符串计算的,最好都多预留一个,比较好算。
int minDistance(string word1, string word2) {
int m = word1.size();
int n = word2.size();
vector<vector<int>> dp(m+1, vector<int>(n+1, 0));
// 初始化
for (int i = 1; i <= m; i++) dp[i][0] = i;
for (int j = 1; j <= n; j++) dp[0][j] = j;
// 计算 dp 数组
for (int i = 1; i <= m; i++) {
for (int j = 1; j <= n; j++) {
if (word1[i-1] == word2[j-1]) {
dp[i][j] = dp[i-1][j-1];
} else {
int m1 = min(dp[i][j-1], dp[i-1][j]);
dp[i][j] = min(m1, dp[i-1][j-1]) + 1;
}
}
}
return dp[m][n];
}





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