LeetCode题解——Edit Distance

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

分析：本题用 动态规划的思想解答。用二维数组p[i][[j]表示word1[1...i]变为word2[1...j]需要经过最少的步数。若已知p[i][j]要p[i+1][j+1]对应3种操作，有三种转化方式
1. 修改一个字符。
当word1[i+1] != word2[j+1]时，需要修改字符，而word1[1…i]变换到word2[1...j]已经求得为p[i][j], 因此修改字符时，p[i+1][j+1] = p[i][j] + (word1[i+1] == word2[j+1]? 0:1)
2. 插入一个字符。
插入一个字符是在将word1[1...i+1]变换到word2[1...j]的基础上，插入一个字符使得word1[1...i+1]变换到word2[1....j+1]，因此p[i+1][j+1] = p[i+1][j]+1;
3. 删除一个字符
需要删除时，说明字符word1[i+1]是多余的，删除后的字符串结果与将word1[1...i]变换到word2[1...j+1]相同，只是多了一步操作。因此p[i+1][j+1] = p[i][j+1]+1

p[i+1][j+1]取这三种变换的最小值。

所以
  p[i][j] = min(p[i-1][j-1] +(word1[i] == word2[j] ? 0 :1), min(p[i][j-1] +1, p[i-1][j] +1) )

初始化：
p[0][j] 表示将空字符串转化成word2[1...j]，因此 p[0][j] = j
p[i][0] 表示将word1[1…i]转化成空字符串，因此p[i][0] = i

算法复杂度O(mn)
代码：

public class Edit_Distance {
	public int minDistance(String word1, String word2) {
		
		int p[][] = new int[word1.length()+1][word2.length()+1];
		for(int i=0;i<p[0].length;++i)
			p[0][i] = i;
		for(int i=0;i<p.length;++i)
			p[i][0] = i;
		for(int i=1;i<=word1.length();++i){
			for(int j=1;j<=word2.length();++j){
				p[i][j] = Math.min(p[i][j-1]+1, Math.min(p[i-1][j]+1, p[i-1][j-1]+(word1.charAt(i-1)== word2.charAt(j-1)? 0: 1)));
			}
		}
		return  p[word1.length()][word2.length()];
	}
	
}