算法作業_26(2017.6.6第十六週)

583. Delete Operation for Two Strings

Given two words word1 and word2, find the minimum number of steps required to make word1 and word2 the same, where in each step you can delete one character in either string.

Example 1:

Input: "sea", "eat"
Output: 2
Explanation: You need one step to make "sea" to "ea" and another step to make "eat" to "ea".

思路：還是動態規劃問題，求最長子序列，思路同上一篇：

public class Solution {
    public int minDistance(String word1, String word2) {
        int len1 = word1.length();
        int len2 = word2.length();
        if(len1==0)
            return len2;
        if(len2==0)
            return len1;

        int[][] dp = new int[len1+1][len2+1];
        
        for(int i = 0 ; i<=len1;i++){
            for(int j =0;j<=len2;j++){
                if(i==0||j==0){
                    dp[i][j] = 0;
                }else if(word1.charAt(i-1)==word2.charAt(j-1)){
                    dp[i][j] = dp[i-1][j-1] + 1;
                }else{
                    dp[i][j] = Math.max(dp[i-1][j] , dp[i][j-1]);
                }
            }
        }
        int maxLength = dp[len1][len2];
        return  len1-maxLength+len2-maxLength;
         
    }
}

class Solution {
public:
    int minDistance(string word1, string word2) {
        int len1 = word1.size();
        int len2 = word2.size();
        
        vector<vector<int>> dp (len1+1,vector<int>(len2+1));
        
        for(int i = 0;i<=len1;i++){
            for(int j = 0;j<=len2;j++){
                if(i==0 || j==0){
                    dp[i][j] = 0;
                }else if(word1[i-1] == word2[j-1]){
                    dp[i][j] = dp[i-1][j-1] +1;
                }else{
                    dp[i][j] = max(dp[i-1][j],dp[i][j-1]);
                }
            }
        }
        
        return len1+len2-2*dp[len1][len2];
    }
};