# Problem: Longest Common Subsequence (LintCode)

> [http://www.lintcode.com/en/problem/longest-common-subsequence/](http://www.lintcode.com/en/problem/longest-common-subsequence/)

## 思路

![](https://1241747088-files.gitbook.io/~/files/v0/b/gitbook-legacy-files/o/assets%2F-Lpv9LvBSlFaukf_ALqh%2F-Lpv9NPw3Ji1X5Vk8CES%2F-Lpv9xhMn1IVkeCHV315%2FLSC_1.jpg?generation=1569729545584316\&alt=media)

```java
public class Solution {
    /**
     * @param A, B: Two strings.
     * @return: The length of longest common subsequence of A and B.
     */
    public int longestCommonSubsequence(String A, String B) {
        int n = A.length();
        int m = B.length();

        int[][] f = new int[n + 1][m + 1];
        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] = f[i - 1][j - 1] + 1;
                }
            }
        }

        return f[n][m];
    }
}
```

## 易错点

1.  数组的初始化

    ```java
    int[][] f = new int[n + 1][m + 1];
    ```

    只要数组初始化好了，**每个元素的值已经为0了**。这道题看似没有初始化，实际上因为数组初始化后， 已经吧`f[i][0]`和`f[0][j]`初始化为0，所以后面直接进行了操作。
2.  两种条件

    `f[i][j] = f[i - 1][j - 1] + 1;`这是A\[i]等于B\[j]下的条件；如果不等的话，那就是`f[i][j] = Math.max(f[i - 1][j], f[i][j - 1]);`
