Leetcode72

看起来比较棘手的一道题（列DP方程还是要大胆猜想。。）

DP方程该怎么列呢？

dp[i][j]表示字符串a[0....i-1]转化为b[0....j-1]的最少距离

转移方程分三种情况考虑 分别对应三中操作

因为只需要三个值就可以更新dp[i][j] 我们可以把空间复杂度降低到O（n）

1. Replace word1[i - 1] by word2[j - 1] (dp[i][j] = dp[i - 1][j - 1] + 1 (for replacement));
2. Delete word1[i - 1] and word1[0..i - 2] = word2[0..j - 1] (dp[i][j] = dp[i - 1][j] + 1 (for deletion));
3. Insert word2[j - 1] to word1[0..i - 1] and word1[0..i - 1] + word2[j - 1] = word2[0..j - 1] (dp[i][j] = dp[i][j - 1] + 1 (for insertion)).

   class Solution { 
   public:
       int minDistance(string word1, string word2) {
           int m = word1.length(), n = word2.length();
           vector<int> cur(m + 1, 0);
           for (int i = 1; i <= m; i++)
               cur[i] = i;
           for (int j = 1; j <= n; j++) {
               int pre = cur[0];
               cur[0] = j;
               for (int i = 1; i <= m; i++) {
                   int temp = cur[i];
                   if (word1[i - 1] == word2[j - 1])
                       cur[i] = pre;
                   else cur[i] = min(pre + 1, min(cur[i] + 1, cur[i - 1] + 1));
                   pre = temp;
               }
           }
           return cur[m]; 
       }
   }; 
   

   　　

第十八周 Leetcode 72. Edit Distance(HARD) O(N^2)DP