leetcode 72.

生命不息，奋斗不止！

@author stormma
@date 2017/10/21

题目
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

思路分析
我们先比较word1和word2的最后一位，如果最后一位相同的话，那问题就转换成了word1不包括最后一位的字符串和word2不包括最后一位的字符串直接的转化，同理如果不相同，当前操作数+1再加word1的前i位–>word2的前j-1位，word1的前i-1转换成word2的前j位，word1的前i-1位转换成word2的前j-1位的最小操作数。其实就是一个dp的问题，难度系数：4星，状态比较多，难以考虑周全。

代码实现

public class Question72 {    // 记忆化搜索    static class Solution {        private int solve(String word1, String word2, int pos1, int pos2, int[][] memo) {            if (pos1 == 0) {                return memo[pos1][pos2] = pos2;            }            if (pos2 == 0) {                return memo[pos1][pos2] = pos1;            }            if (word1.charAt(pos1 - 1) == word2.charAt(pos2 - 1)) {                if (memo[pos1 - 1][pos2 - 1] != 0) {                    return memo[pos1 - 1][pos2 - 1];                }                return memo[pos1][pos2] = solve(word1, word2, pos1 - 1, pos2 - 1, memo);            } else {                if (memo[pos1 - 1][pos2] == 0) {                    memo[pos1 - 1][pos2] = solve(word1, word2, pos1 - 1, pos2, memo);                }                if (memo[pos1][pos2 - 1] == 0) {                    memo[pos1][pos2 - 1] = solve(word1, word2, pos1, pos2 - 1, memo);                }                if (memo[pos1 - 1][pos2 - 1] == 0) {                    memo[pos1 - 1][pos2 - 1] = solve(word1, word2, pos1 - 1, pos2 - 1, memo);                }                return 1 + Math.min(Math.min(memo[pos1 - 1][pos2], memo[pos1][pos2 - 1]), memo[pos1 - 1][pos2 - 1]);            }        }        public int minDistance(String word1, String word2) {            int[][] memo = new int[word1.length() + 1][word2.length() + 1];            return solve(word1, word2, word1.length(), word2.length(), memo);        }    }    // dp    static class Solution2 {        public int minDistance(String word1, String word2) {            // dp[i][j] 表示word1的前i个字符转换成word2的前j个字符需要操作次数            int[][] dp = new int[word1.length() + 1][word2.length() + 1];            for (int i = 0; i <= word1.length(); i++) {                dp[i][0] = i;            }            for (int j = 0; j <= word2.length(); j++) {                dp[0][j] = j;            }            for (int i = 1; i <= word1.length(); i++) {                for (int j = 1; j <= word2.length(); j++) {                    int c = (word1.charAt(i - 1) == word2.charAt(j - 1)) ? 0 : 1;                    dp[i][j] = Math.min(dp[i - 1][j - 1] + c, 1 + Math.min(dp[i - 1][j], dp[i][j - 1]));                }            }            return dp[word1.length()][word2.length()];        }    }}