[LeetCode]Edit Distance

最新推荐文章于 2020-06-22 10:16:26 发布

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本文提供了一种使用动态规划解决编辑距离问题的C++实现方案。通过构建二维动态规划矩阵，计算两个字符串之间的最小编辑距离，即进行插入、删除或替换操作使两字符串相等所需的最少步骤。

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class Solution {
//DP
public:
	int minDistance(string word1, string word2) {
		// Start typing your C/C++ solution below
		// DO NOT write int main() function
		int len1 = word1.size();
		int len2 = word2.size();
		if(len1 == 0 || len2 == 0) return max(len1, len2);
		vector<vector<int>> f(len1+1, vector<int>(len2+1));
		//init
		for(int i = 0; i <= len1; ++i) 
			f[i][0] = i;
		for(int j = 0; j <= len2; ++j) 
			f[0][j] = j;
		//transform
		for (int i = 1; i <= len1; ++i)
		{
			for (int j = 1; j <= len2; ++j)
			{
				if(word1[i-1] == word2[j-1])
					f[i][j] = f[i-1][j-1];
				else
				{
					int tmp = min(f[i-1][j], f[i][j-1]);
					f[i][j] = min(tmp, f[i-1][j-1])+1;
				}
				
			}
		}

		return f[len1][len2];
	}
};

second time

class Solution {
public:
    int minDistance(string word1, string word2) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int n1 = word1.size();
        int n2 = word2.size();
        vector<vector<int> > f(n1+1, vector<int>(n2+1, INT_MAX));
        //initialize
        for(int i = 0; i <= n1; ++i) f[i][0] = i;
        for(int j = 0; j <= n2; ++j) f[0][j] = j;
        //dp
        for(int i = 1; i <= n1; ++i)
        {
            for(int j = 1; j <= n2; ++j)
            {
                if(word1[i-1] == word2[j-1]) f[i][j] = f[i-1][j-1];
                int tmp = INT_MAX;
                tmp = min(tmp, f[i-1][j]);
                tmp = min(tmp, f[i][j-1]);
                tmp = min(tmp, f[i-1][j-1]);
                
                f[i][j] = min( f[i][j], tmp+1 );
            }
        }
        return f[n1][n2];
    }
};