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”.

Note:

The length of given words won't exceed 500.Characters in given words can only be lower-case letters.

此类型的题目属于动态规划；
建立一个二维数组，对于dp[i][j]，表示word1第i个元素匹配word2第j个元素所需要删除的次数；
如果word1[i] == word2[j]，则表示两个字符是相同的，则dp[i][j] = dp[i-1][j-1]；
如果word1[i] ！= word2[j]，则表示两个字符是不相同的，则dp[i][j] = min( dp[i-1][j], dp[i][j-1] ) + 1；

代码如下：

int minDistance1(string word1, string word2) {    int m = word1.size(), n = word2.size();    vector<vector<int>> dp(m + 1, vector<int>(n + 1, 0));    for (int i = 1; i <= m; i++)dp[i][0] = i;    for (int j = 1; j <= n; j++)dp[0][j] = j;    for (int i = 1; i <= m; i++){        for (int j = 1; j <= n; j++){            if (word1[i - 1] == word2[j - 1])dp[i][j] = dp[i - 1][j - 1];            else dp[i][j] = min(dp[i - 1][j], dp[i][j - 1]) + 1;        }    }    return dp[m][n];}