leetcode 72.

生命不息，奮鬥不止！

@author stormma
@date 2017/10/21

題目
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
我們先比較word1和word2的最後一位，如果最後一位相同的話，那問題就轉換成了word1不包括最後一位的字符串和word2不包括最後一位的字符串直接的轉化，同理如果不相同，當前操作數+1再加word1的前i位–>word2的前j-1位，word1的前i-1轉換成word2的前j位，word1的前i-1位轉換成word2的前j-1位的最小操作數。其實就是一個dp的問題，難度係數：4星，狀態比較多，難以考慮周全。

代碼實現

public class Question72 {
    // 記憶化搜索
    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個字符轉換成word2的前j個字符需要操作次數
            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()];
        }
    }
}