LeetCode-198. House Robber (JAVA)寻找数组不相邻组合最大值DP

198. House Robber

You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed, the only constraint stopping you from robbing each of them is that adjacent houses have security system connected and it will automatically contact the police if two adjacent houses were broken into on the same night.

Given a list of non-negative integers representing the amount of money of each house, determine the maximum amount of money you can rob tonight without alerting the police.

这道题的本质相当于在一列数组中取出一个或多个不相邻数，使其和最大。

这题的考点是动态规划。
动态规划一定要找递推公式！！！

1. 对这题来说，对每一家房子，在其前一家房子偷不偷的前提下，有两种可能的情况。
2. 前一家房子被偷了，它就不能再偷了。前一家房子没被偷，它可以被偷也可以选择不偷。
3. 可看出，每一个子问题都依赖于前一个子问题，同时每一个子问题都会产生至少一种情况（之多两种情况）状态转移方程：

dp[0] = num[0] （当i=0时）
dp[1] = max(num[0], num[1]) （当i=1时）
dp[i] = max(num[i] + dp[i - 2], dp[i - 1])   （当i !=0 and i != 1时）

	public int rob(int[] nums) {
		if (nums == null)
			return 0;
		int n = nums.length;
		if (n == 0)
			return 0;
		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 - 2]
					+ nums[i], dp[i - 1]);
		return dp[n - 1];
	}

空间优化

用本身数组代替dp数组。

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