718. Maximum Length of Repeated Subarray

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DP

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class Solution {
    public int findLength(int[] nums1, int[] nums2) {
        int m = nums1.length, n = nums2.length;

        int maxLength = 0;

        // 定义 dp[i][j] 为 nums1[0, i - 1] 与 nums[0, j - 1] 的最长重复后缀子数组的长度
        int[][] dp = new int[m + 1][n + 1];
        for (int i = 1; i <= m; i++) {
            for (int j = 1; j <= n; j++) {
                if (nums1[i - 1] == nums2[j - 1]) { // 注意索引映射不要出错
                    dp[i][j] = dp[i - 1][j - 1] + 1;
                    maxLength = Math.max(maxLength, dp[i][j]);
                } else {
                    dp[i][j] = 0;
                }
            }
        }

        return maxLength;
    }
}

因为 dp[i][j] 仅依赖 dp[i - 1][j - 1]，所以可以降为一维数组：

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class Solution {
    public int findLength(int[] nums1, int[] nums2) {
        int[] dp = new int[nums2.length + 1]; // depends on dp[j - 1]
        int maxLength = 0;
        for (int i = 1; i <= nums1.length; i++) {
            for (int j = nums2.length; j >= 1; j--) { // 因为 dp[j] 依赖 dp[j - 1], 所以从右往左遍历
                if (nums1[i - 1] == nums2[j - 1]) {
                    dp[j] = dp[j - 1] + 1;
                }

                maxLength = Math.max(maxLength, dp[j]);
            }
        }

        return maxLength;
    }
}

Sliding Window

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class Solution {
    public int findLength(int[] nums1, int[] nums2) {
        // nums1 = [1, 2, 3, 2, 1], nums2 = [3, 2, 1, 4, 7], 固定其中一个数组，移动另一个数组

        int maxLength = 0;

        // nums1:             [1, 2, 3, 2, 1]
        // nums2: [3, 2, 1, 4, 7]
        for (int j = nums2.length - 1; j >= 0; j--) {
            int length = findCommonLength(nums1, 0, nums2, j);
            maxLength = Math.max(maxLength, length);
        }

        // nums1: [1, 2, 3, 2, 1]
        // nums2:    [3, 2, 1, 4, 7]
        for (int i = 1; i < nums1.length; i++) {
            int length = findCommonLength(nums1, i, nums2, 0);
            maxLength = Math.max(maxLength, length);
        }

        return maxLength;
    }

    private int findCommonLength(int[] nums1, int i, int[] nums2, int j) {
        int maxLength = 0;

        int length = 0;
        while (i < nums1.length && j < nums2.length) {
            if (nums1[i] == nums2[j]) {
                length++;
                maxLength = Math.max(maxLength, length);
            } else {
                length = 0;
            }
            i++;
            j++;
        }

        return maxLength;
    }
}

References

718. Maximum Length of Repeated Subarray