算法期中练习——1006. 最长公共子串

算法期中练习——1006. 最长公共子串

Description:

给定两个字符串x = x­1x­2…x­n和y = y­1y­2…ym, 请找出x和y的最长公共子串的长度，也就是求出一个最大的k，使得存在下标i和j有x­ix­i+1…x­i+k-1 = yjy­j+1…y­j+k-1.
x和y只含有小写字母，长度均在1和1000之间.

Example:

例1：x = “abcd”, y = “cdef”，返回值为2.

例2：x = “abcabc”, y = “xyz”，返回值为0.

例3：x = “introduction”, y = “introductive”，返回值为10.

请为下面的Solution类实现解决上述问题的函数longestSubstring，函数的参数x和y为给出的两个单词，返回值为最长公共子串的长度.

class Solution {
public:
    int longestSubstring(string x, string y) {
       
       }
};

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代码如下：

class Solution {
public:
    int longestSubstring(string x, string y) {
        int a = x.length();
        int b = y.length();
        int dp[a + 1][b + 1];
        int max = 0;
        for (int j = 0; j < b; j++) dp[0][j] = 0;
        for (int i = 0; i < a; i++) dp[i][0] = 0;

        for (int i = 1; i <= a; i++) {
            for (int j = 1; j <= b; j++) {
                if (x[i-1] == y[j-1]) {
                    dp[i][j] = dp[i - 1][j - 1] + 1;
                    if (max < dp[i][j]) max = dp[i][j];
                }
                else {
                    dp[i][j] = 0;
                }
            }
        }
        return max;
    }
};
// another solution, but is the same
class Solution {
private:
    vector<vector<int>> f;
public:
    int longestSubstring(string x, string y) {
        int n, m, i, j;
        n = x.length();
        m = y.length();
        f.resize(n + 1);
        for (i = 0; i <= n; i++) f[i].resize(m + 1);
        int res = 0;
        for (i = 0; i <= n; i++) {
            for (j = 0; j <= m; j++) {
                if (i == 0 || j == 0 || x[i - 1] != y[j - 1]) f[i][j] = 0;
                else f[i][j] = f[i - 1][j - 1] + 1;
                if (f[i][j] > res) res = f[i][j];
            }
        }
        return res;
    }
};