LC.P1035[不相交的线]

方法一：动态规划

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class Solution {
    public int maxUncrossedLines(int[] nums1, int[] nums2) {
        int n = nums1.length, m = nums2.length;
        int[][] f = new int[n + 1][m + 1];
        for (int i = 1; i <= n; ++i) {
            for (int j = 1; j <= m; ++j) {
                if (nums1[i - 1] == nums2[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]);
                }
            }
        }
        return f[n][m];
    }
}
  • 时间复杂度：$O(n \times m)$
  • 空间复杂度：$O(n \times m)$