[LeetCode]198. House Robber I&II

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https://leetcode.com/problems/house-robber/

显然空间复杂度O(1)的DP问题。rob和notRob记录当前位置抢或者不抢所能获得的最大收益。当前未抢 = max(上次抢了，上次未抢)；当前抢了 = 上次未抢 + 当前价值

public class Solution {    public int rob(int[] nums) {        if (nums == null) {            return 0;        }        int rob = 0;        int notRob = 0;        for (int n : nums) {            int curNotRob = notRob;            notRob = Math.max(notRob, rob);            rob = curNotRob + n;        }        return Math.max(rob, notRob);    }}

https://leetcode.com/problems/house-robber-ii/

如果数组成环。

那么首尾两个只能最多取一个，因此结果为1 ~ len - 1 和 0 ~ len - 2里面较大的那个

public class Solution {    public int rob(int[] nums) {        if (nums == null || nums.length == 0) {            return 0;        }        return Math.max(rob(nums, 0, nums.length - 2), rob(nums, 1, nums.length - 1));    }    private int rob(int[] nums, int beg, int end) {        if (nums.length == 1) {            return nums[0];        }        int notRob = 0;        int rob = 0;        for (int i = beg; i <= end; i++) {            int curNotRob = Math.max(notRob, rob);            rob = notRob + nums[i];            notRob = curNotRob;        }        return Math.max(rob, notRob);    }}

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