LeetCode link

first thought

  • Using DP. Build a int[][] to save the maximum between i to j.

solution

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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][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; k++) {
            for (int i = 0; i + k < n; i++) {
                int j = i + k;
                dp[i][j] = Math.max(dp[i][j - 1], dp[i][j - 2] + nums[j]);
            }
        }
        return dp[0][n - 1];
    }
}

promote

There’s no need to build a 2d-array, 1d is enough.

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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];
        dp[1] = Math.max(nums[0], nums[1]);
        for (int i = 2; i < n; i++) {
            dp[i] = Math.max(dp[i - 1], dp[i - 2] + nums[i]);
        }
        return dp[n - 1];
    }
}