Given two words word1 and word2, find the minimum number of steps required to convert word1 to word2. (each operation is counted as 1 step.)
You have the following 3 operations permitted on a word:
a) Insert a character
b) Delete a character
c) Replace a character
状态方程 f[i][j] 表示word1的前i个字符转换成word2的前jg字符所需要的最小次数。
两种情况:
if(word1.charAt(i - 1) == word2.charAt(j - 1)) {
f[i][j] = Math.min(f[i - 1][j - 1], Math.min(f[i][j - 1] + 1, f[i - 1][j] + 1));
} else {
f[i][j] = Math.min(f[i - 1][j - 1] + 1, Math.min(f[i][j - 1] + 1, f[i - 1][j] + 1));
}
public class Solution { public int minDistance(String word1, String word2) { if(word1 == null && word2 == null) { return 0; } if(word1 == null) { return word2.length(); } if(word2 == null) { return word1.length(); } int m = word1.length(); int n = word2.length(); int[][] f = new int[m + 1][n + 1]; for(int i = 1; i <= m; i++) { f[i][0] = i; } for(int j = 1; j <= n; j++) { f[0][j] = j; } for(int i = 1; i <= m; i++) { for(int j = 1; j <= n; j++) { if(word1.charAt(i - 1) == word2.charAt(j - 1)) { f[i][j] = Math.min(f[i - 1][j - 1], Math.min(f[i][j - 1] + 1, f[i - 1][j] + 1)); } else { f[i][j] = Math.min(f[i - 1][j - 1] + 1, Math.min(f[i][j - 1] + 1, f[i - 1][j] + 1)); } } } return f[m][n]; } }