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ref: https://discuss.leetcode.com/topic/3136/my-o-mn-time-and-o-n-space-solution-using-dp-with-explanation/2
如果用dp[i][j]表示word1里[0,i]长的子串和word2[0,j]的最小编辑距离，那么状态转移方程应该这样构成：
dp[i][j] = dp[i-1][j-1], if word1[i] == word2[j] dp[i][j] = max{dp[i-1][j], dp[i][j-1], dp[i-1][j-1]} + 1, otherwise
如果word1[i] == word2[j]的时候不需要解释，
如果不相等的时候可能有三种操作，使两个字符串对齐：
1. 替换字母：把word1[i] 替换成word2[j]，那么是dp[i-1][j-1]+1
2. 在Word1[i]后面强行插入一个word2[j]. 所以可能是dp[i][j-1]
3. 直接把word1[i]删了，dp[i-1][j]
所以用二维矩阵表示很容易：
1 public int minDistance(String word1, String word2) { 2 int len1 = word1.length(); 3 int len2 = word2.length(); 4 5 //distance[i][j] is the distance converse word1(1~ith) to word2(1~jth) 6 int[][] distance = new int[len1 + 1][len2 + 1]; 7 for (int j = 0; j <= len2; j++) 8 {distance[0][j] = j;} //delete all characters in word2 9 for (int i = 0; i <= len1; i++) 10 {distance[i][0] = i;} 11 12 for (int i = 1; i <= len1; i++) { 13 for (int j = 1; j <= len2; j++) { 14 if (word1.charAt(i - 1) == word2.charAt(j - 1)) { //ith & jth 15 distance[i][j] = distance[i - 1][j - 1]; 16 } else { 17 distance[i][j] = Math.min(Math.min(distance[i][j - 1], distance[i - 1][j]), distance[i - 1][j - 1]) + 1; 18 } 19 } 20 } 21 return distance[len1][len2]; 22 }
但是，可以像uniquePath一样把二维压成一维
pre表示上同一行前一列的值，f[j-1]表示f[i-1][j-1]（因为pre没有更新进去），f[j]表示同一列上一行
1 public int minDistance(String word1, String word2) { 2 if(word1 == null || word2 == null) { 3 return -1; 4 } 5 int len1 = word1.length(); 6 int len2 = word2.length(); 7 int[] distance = new int[len2 + 1]; 8 for(int i = 0; i <= len2; i++) { 9 distance[i] = i; 10 } 11 for(int i = 1; i <= len1; i++) { 12 int pre = i; 13 for(int j = 1; j <= len2; j++) { 14 int cur = 0; 15 if(word1.charAt(i - 1) == word2.charAt(j - 1)) { 16 cur = distance[j - 1]; 17 } else { 18 cur = Math.min(pre, Math.min(distance[j], distance[j - 1])) + 1; 19 } 20 distance[j - 1] = pre; 21 pre = cur; 22 } 23 distance[len2] = pre; 24 } 25 return distance[len2]; 26 }
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原文地址：http://www.cnblogs.com/warmland/p/5971909.html
