/** * dp[r][c] indicates the minimum number of operations to convert word1[0,r] to word2[0,c] * boundary conditions: * dp[r][0] = r * dp[0][c] = c * <p> * if word1[r] == word2[c]:  dp[r][c] = dp[r-1][c-1] * else:  dp[r][c] = min(dp[r-1][c-1]+1, dp[r-1][c]+1, dp[r][c-1]+1) * <p> * dp[r-1][c-1]+1 means "replace the last character of word1 with the last character of word2" * dp[r-1][c]+1 means "delete the last character from word1": *    dp[r-1][c] already matches. The extra word1[r] is useless, so we remove it * dp[r][c-1]+1 means "add the last character of word2 to the end of word1": *    dp[r][c-1] already matches. In order to match word2[c], we need to add it to the end of word1 */class Solution {  public int minDistance(String word1, String word2) {    int ROW = word1.length();    int COL = word2.length();    int[][] dp = new int[ROW + 1][COL + 1];    for (int r = 1; r <= ROW; ++r)      dp[r][0] = r;    for (int c = 1; c <= COL; ++c)      dp[0][c] = c;    for (int r = 1; r <= ROW; ++r) {      for (int c = 1; c <= COL; ++c) {        if (word1.charAt(r - 1) == word2.charAt(c - 1))          dp[r][c] = dp[r - 1][c - 1];        else          dp[r][c] = Math.min(dp[r - 1][c - 1] + 1, Math.min(dp[r][c - 1] + 1, dp[r - 1][c] + 1));      }    }    return dp[ROW][COL];  }}

161. One Edit Distance