public int minDistance(String word1, String word2) {if (word1 == null && word2 == null)return 0;if (word1 == null || word1.isEmpty())return word2.length();if (word2 == null || word2.isEmpty())return word1.length();int m = word1.length();int n = word2.length();// Cost[i][j]表示words1.substr(0, i)到 words2.substr(0, j) 的最短编辑距离int[][] Cost= new int[m+1][n+1];// 初始化边界情况for (int i = 0; i <= m; ++i)Cost[i][0] = i;for (int j = 0; j <= n; ++j)Cost[0][j] = j;// 由A[0…i]到B[0…j]的最短距离分为三种情况for (int i = 1; i <= m; ++i) {for (int j = 1; j <= n; ++j) {int insertBoth = Cost[i-1][j-1] + replaceCost(word1, word2,0,0);int insertA = Cost[i-1][j] + 1;int insertB = Cost[i][j-1] + 1;Cost[i][j] = Math.min(Math.min(insertA, insertB), insertBoth);}}return Cost[m][n]; }private int replaceCost(String word1, String word2, int i1, int i2) {return word1.charAt(i1) == word2.charAt(i2) ? 0 : 1;}参考：
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