LintCode-编辑距离

给出两个单词word1和word2，计算出将word1 转换为word2的最少操作次数。

你总共三种操作方法：

插入一个字符
删除一个字符
替换一个字符

样例
给出 work1="mart" 和 work2="karma"

返回 3

分析:minSteps[i][j]表示word1的前i个字符改为word2的前j个字符的最少操作数，因此有转移方程
minSteps[i][j] =
{
minSteps[i - 1][j - 1];(word1[i] == wrod2[j])
min(minSteps[i - 1][j - 1],minSteps[i][j - 1],minSteps[i - 1][j]);(word1[i] != wrod2[j])
}

public class Solution {
    public int minDistance(String word1, String word2) {
        if(null == word1 || null == word2)
            return 0;
        
        int[][] minSteps = new int[word1.length() + 1][word2.length() + 1];
        for(int i = 0;i <= word1.length();i++)
        {
            minSteps[i][0] = i;
        }
        
        for(int i = 0;i <= word2.length();i++)
        {
            minSteps[0][i] = i;
        }
        
        for(int i = 1;i <= word1.length();i++)
        {
            for(int j = 1;j <= word2.length();j++)
            {
                if(word1.charAt(i - 1) == word2.charAt(j - 1))
                {
                    minSteps[i][j] = minSteps[i - 1][j - 1];
                }
                else
                {
                    minSteps[i][j] = Math.min(minSteps[i - 1][j - 1], Math.min(minSteps[i - 1][j], minSteps[i][j - 1])) + 1;
                }
            }
        }
        return minSteps[word1.length()][word2.length()];
    }
};