Edit Distance 解题报告

题目链接
这是道比较经典的动态规划题。
状态转移方程:

对于字符串x,y,编辑距离E(i,j)为E(i,j) = min(1 + E(i-1,j), 1 + E(i,j-1), diff(i,j) + E(i-1,j-1))diff(i,j) = if x[i] == y[j] then 0 else 1

初始状态：

E(0,i) = E(j,0) = 0

伪代码是

for i = 0,1,2,...,m:    E(i,0) = ifor j = 1,2,...,m:    E(0,j) = jfor i = 1,2,...,m:    for j = 1,2,...,n:        diff(i,j) = if x[i] == y[j] then 0 else 1        E(i,j) = min(1 + E(i-1,j), 1 + E(i,j-1), diff(i,j) + E(i-1,j-1))容易看出复杂度是O(mn)的

c++代码实现:

class Solution {public:    int minDistance(string word1, string word2) {        int r = word1.length() + 1;        int c = word2.length() + 1;        int **E = new int*[r];        for(int i = 0; i < r; i++){            E[i] = new int[c];            for(int j = 0; j < c; j++){                E[i][j] = 0;            }        }        for(int i = 0; i < r; i++){            E[i][0] = i;        }        for(int j = 0; j < c; j++){            E[0][j] = j;        }        for(int i = 1; i < r; i++){            for(int j = 1; j < c; j++){                E[i][j] = min(min(1+E[i-1][j],1+E[i][j-1]), int(!(word1[i-1] == word2[j-1])) + E[i-1][j-1]);            }        }        return E[r-1][c-1];    }};

因为字符数组下标从0开始，其中转化走了一些弯路。
提交结果如下：