算法作业_26(2017.6.6第十六周)

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：还是动态规划问题，求最长子序列，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];
}
};