题目链接
题目描述
实现代码
动态规划思想:记dp[i][j]表示从word1前i个字符到word2的前j个字符的编辑距离,有两种情况:
- 当word1[i] == word2[j]时:dp[i][j] = dp[i-1][j-1]
- 当word1[i] ≠ word2[j]时:需要进行插入、删除或者替换操作,从三种操作的结果中取最小
三种操作的变换方式(变换的都是word1):
- 插入:dp[i][j] = dp[i][j-1]
- 删除:dp[i][j] = dp[i-1][j]
- 替换:dp[i][j] = dp[i-1][j-1]
初始化条件:
- 对i = 0时,dp[i][j] = j
- 对j = 0时,dp[i][j] = i
实现代码:
class Solution {
public int minDistance(String word1, String word2) {
int n = word1.length();
int m = word2.length();
// 有一个字符串为空串
if (n * m == 0) {
return n + m;
}
// DP 数组
int[][] D = new int[n + 1][m + 1];
// 边界状态初始化
for (int i = 0; i < n + 1; i++) {
D[i][0] = i;
}
for (int j = 0; j < m + 1; j++) {
D[0][j] = j;
}
// 计算所有 DP 值
for (int i = 1; i < n + 1; i++) {
for (int j = 1; j < m + 1; j++) {
int left = D[i - 1][j] + 1;
int down = D[i][j - 1] + 1;
int left_down = D[i - 1][j - 1];
if (word1.charAt(i - 1) != word2.charAt(j - 1)) {
left_down += 1;
}
D[i][j] = Math.min(left, Math.min(down, left_down));
}
}
return D[n][m];
}
}