Given two strings, find the longest common substring.

Return the length of it.

 Notice

The characters in substring should occur continuously in original string. This is different with subsequence.

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Example
Given A = "ABCD", B = "CBCE", return 2
 
.

Challenge 
O(n x m) time and memory.

區別於 Longest Common Subsequence 這是累加求subString 的

狀態設為遍歷到當前的i, j時的結果值, 但是卻是不對, 是因為求得是累加的值,

for (int i = 1; i <= n; i++) {
           for (int j = 1; j <= m; j++) {
                f[i][j] = Math.max(f[i - 1][j], f[i][j - 1]);
                if (A.charAt(i - 1) == B.charAt(j - 1)) {
                    f[i][j] = Math.max(f[i - 1][j - 1] + 1, f[i][j]);
                } 
            
            }
        }

和這樣寫無差異, 都是求的累加和.

if (A.charAt(i - 1) == B.charAt(j - 1)) {
                    f[i][j] = f[i - 1][j - 1] + 1;
                } else {
                    f[i][j] = Math.max(f[i - 1][j], f[i][j - 1]);
                }

因此要改下狀態(根據結果來設定狀態需要那幾個變量表示, 這個狀態跟), : 匹配到的字符是最後一個字符的結果值(作為局部變量 ) + 全局變量

常見狀態:

1遍歷到當前字符時的最優解,

2 遍歷到當前字符,字符如果是題意要求的string的局部解

方程: 分情況討論, 根據題意與當前字符的形態和上個狀態做轉化

public int longestCommonSubstring(String A, String B) {
        // write your code here
        //state
        int n = A.length();
        int m = B.length();
        int[][] f = new int[n + 1][m + 1];
        //initialize
        for (int i = 0; i <= n; i++) {
            f[i][0] = 0;
        }
        for (int i = 0; i <= m; i++) {
            f[0][i] = 0;
        }
        int ans = 0;
        //function
        for (int i = 1; i <= n; i++) {
           for (int j = 1; j <= m; j++) {
            if (A.charAt(i - 1) == B.charAt(j - 1)) {
                f[i][j] = f[i - 1][j - 1] + 1;
            } else {
                f[i][j] = 0;
            }
            ans = Math.max(f[i][j], ans);
            
        }
        }
        return ans;
    }

Longest Common Substring