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Given two words word1 and word2, find the minimum number of steps required to convert word1 to word2. (each operation is counted as 1 step.)

You have the following 3 operations permitted on a word:

a) Insert a character
b) Delete a character
c) Replace a character

思路

d[i][j]表示word1的前i个字符到word2的前j个字符的距离

一、word1[i] == word2[j] 时， d[i][j] = d[i-1][j-1]

二、word1[i] != word2[j] 时，取下面三种情况的最小者

1. d[i-1][j] + 1 （添加）

2. d[i][j-1] + 1 （删除）

3. d[i-1][j-1] + 1 （替换）

初始条件：

d[i][0] = i

d[0][j] = j

代码

int minDistance(string word1, string word2) {
	int len1 = word1.size();
	int len2 = word2.size();


	vector<vector<int> > d(len1+1, vector<int>(len2+1, 0));
	for (int i = 0; i <= len1; ++i)
	{
		d[i][0] = i;
	}
	for (int j = 0; j <= len2; ++j)
	{
		d[0][j] = j;
	}
	for (int i = 1; i <= len1; ++i)
	{
		for (int j = 1; j <= len2; ++j)
		{
			if (word1[i-1] == word2[j-1])
				d[i][j] = d[i-1][j-1];
			else
			{
				d[i][j] = min(min(d[i-1][j]+1, d[i][j-1] + 1), d[i-1][j-1]+1);
			}
		}
	}
	return d[len1][len2];
}