213. House Robber II

DP

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class Solution {
    public int rob(int[] nums) {
        if (nums.length == 1) {
            return nums[0];
        }

        int n = nums.length;

        // 定义 dp[i] 为处于第 i + 1 号房屋且 1 号房屋处于 j 状态时偷窃到的最高金额
        // j = 0: 未被偷窃，j = 1: 被偷窃
        int[][] dp = new int[n][2];
        dp[0][0] = 0;
        dp[1][0] = nums[1];

        dp[0][1] = nums[0];
        dp[1][1] = nums[0];

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

        return Math.max(dp[n - 1][0], dp[n - 2][1]);
    }
}

DP

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class Solution {
    public int rob(int[] nums) {
        int n = nums.length;

        if (n == 1) {
            return nums[0];
        }

        return Math.max(rob(nums, 0, n - 2), rob(nums, 1, n - 1));
    }

    private int rob(int[] nums, int start, int end) {
        if (start == end) {
            return nums[start];
        }
        if (start + 1 == end) {
            return Math.max(nums[start], nums[start + 1]);
        }

        // 定义 dp[i] 为截止 nums[i] 能偷盗的最大金额
        int[] dp = new int[end - start + 1];
        dp[0] = nums[start];
        dp[1] = Math.max(nums[start], nums[start + 1]);

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

        return dp[dp.length - 1];
    }
}

References

213. House Robber II
剑指 Offer II 090. 环形房屋偷盗