LeetCode link

Intuition

  • Same as House Robber, just aware when pick the last number, the first one cannot be selected, so the range is 1 to n - 2

solution

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class Solution {
    public int rob(int[] nums) {
        int n = nums.length;
        if (n < 1) {
            return 0;
        }
        if (n == 1) {
            return nums[0];
        }
        int[][] dp = new int[n][n];
        for (int i = 0; i < n; i++) {
            dp[i][i] = nums[i];
        }
        for (int i = 0; i < n - 1; i++) {
            dp[i][i + 1] = Math.max(dp[i][i], dp[i + 1][i + 1]);
        }
        for (int k = 2; k < n - 1; k++) {
            for (int i = 0; i + k < n; i++) {
                int j = i + k;
                if (i == 0 && j == n - 1) {
                    dp[i][j] = Math.max(dp[i][j - 1], dp[i + 1][j - 2] + nums[j]);
                } else {
                    dp[i][j] = Math.max(dp[i][j - 1], dp[i][j - 2] + nums[j]);
                }
            }
        }
        return dp[0][n - 1];
    }
}

problem

WA.

reason

Maybe not that simple, there’s a probability that the first one has not be selected when calculate the dp[0][n - 2]

modification

It’s equals to the problem that Math.max(steal from 0 to n - 2, steal from 1 to n - 1)

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class Solution {
    public int rob(int[] nums) {
        if (nums == null || nums.length < 1) {
            return 0;
        }
        int n = nums.length;
        if (n == 1) {
            return nums[0];
        }
        int[] dp = new int[n];
        dp[0] = nums[0];
        for (int i = 1; i < n - 1; i++) {
            dp[i] = Math.max((i == 1 ? 0 : dp[i - 2]) + nums[i], dp[i - 1]);
        }
        int selected = dp[n - 2];
        dp[1] = nums[1];
        for (int i = 2; i < n; i++) {
            dp[i] = Math.max((i == 2 ? 0 : dp[i - 2]) + nums[i], dp[i - 1]);
        }
        return Math.max(selected, dp[n - 1]);
    }
}