【LeetCode题解】72. Edit Distance （Java）

（一）72. Edit Distance

Given two words word1 and word2, find the minimum number of operations required to convert word1 to word2.

You have the following 3 operations permitted on a word:

Insert a character
Delete a character
Replace a character
Example 1:

Input: word1 = "horse", word2 = "ros"
Output: 3
Explanation: 
horse -> rorse (replace 'h' with 'r')
rorse -> rose (remove 'r')
rose -> ros (remove 'e')

Example 2:

Input: word1 = "intention", word2 = "execution"
Output: 5
Explanation: 
intention -> inention (remove 't')
inention -> enention (replace 'i' with 'e')
enention -> exention (replace 'n' with 'x')
exention -> exection (replace 'n' with 'c')
exection -> execution (insert 'u')

参考：https://leetcode.com/problems/edit-distance/discuss/25849/Java-DP-solution-O(nm)

• 这道题代码和64. Minimum Path Sum非常相似，但是分析起来更难
  • 两者的分析方式相似：都是找到通解，然后再去求特解；并且都是首先初始化一行一列，然后再遍历找最小值
• 对于字符串word1和word2，定义f(i,j)表示实现word1的前i个字符和word2的前j个字符相等需要的最小步骤
  • 如果word1第i个和word2第j个相同，那么不需要进行操作，即f(i,j) = f(i-1,j-1)
  • 如果不同，那么需要进行替换、删除、增加操作则f(i,j) = min(f(i-1,j-1),f(i-1,j),f(i,j-1)) + 1
• 总的来说步骤是1.设置初始值（第一行第一列） 2. 遍历数组，得到每一个点的最小值

package DynamicProgramming.EditDistance;

public class Solution {
	public static int minDistance(String word1, String word2) {
		int len1 = word1.length();
		int len2 = word2.length();
		int[][] cost = new int[len1+1][len2+1];

		for(int i=1; i <= len1; i++){ //初始化第一列
			cost[i][0] = i;
		}

		for(int i=1; i <= len2; i++){ //初始化第一行
			cost[0][i] = i;
		}

		for(int i=1; i <= len1; i++){
			for(int j=1; j <= len2; j++){
				if(word1.charAt(i-1) == word2.charAt(j-1)){
					cost[i][j] = cost[i-1][j-1];
				} else{
					cost[i][j] = findMin(cost[i-1][j-1],cost[i-1][j],cost[i][j-1]);
					cost[i][j] += 1;
				}
			}
		}

		return cost[len1][len2];
	}

	public static int findMin(int a, int b, int c){ //找到三个数中的最小值
		int min = a < b ? a : b;
		min = min < c ? min : c;
		return min;
	}

	public static void main(String[] args){
		String word1 = "horse";
		String word2 = "ros";
		System.out.println(minDistance(word1,word2));
	}
}