README_EN.md

198. House Robber

中文文档

Description

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 systems connected and it will automatically contact the police if two adjacent houses were broken into on the same night.

Given an integer array nums representing the amount of money of each house, return the maximum amount of money you can rob tonight without alerting the police.

 

Example 1:

Input: nums = [1,2,3,1]
Output: 4
Explanation: Rob house 1 (money = 1) and then rob house 3 (money = 3).
Total amount you can rob = 1 + 3 = 4.

Example 2:

Input: nums = [2,7,9,3,1]
Output: 12
Explanation: Rob house 1 (money = 2), rob house 3 (money = 9) and rob house 5 (money = 1).
Total amount you can rob = 2 + 9 + 1 = 12.

 

Constraints:

• 1 <= nums.length <= 100
• 0 <= nums[i] <= 400

Solutions

Python3

class Solution:
    def rob(self, nums: List[int]) -> int:
        def robRange(nums, start, end):
            if end - start == 0:
                return nums[start]
            pre, cur = 0, nums[start]
            for i in range(start + 1, end + 1):
                pre, cur = cur, max(pre + nums[i], cur)
            return cur
        if not nums:
            return 0
        return robRange(nums, 0, len(nums) - 1)

Java

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

    private int robRange(int[] nums, int start, int end) {
        if (end - start == 0) return nums[start];
        int pre = 0;
        int cur = nums[start];
        for (int i = start + 1; i < end + 1; ++i) {
            int t = Math.max(pre + nums[i], cur);
            pre = cur;
            cur = t;
        }
        return cur;
    }
}

C++

class Solution {
public:
    int rob(vector<int>& nums) {
        int n;
        if ((n = nums.size()) == 0) return 0;
        return robRange(nums, 0, n - 1);
    }

private:
    int robRange(vector<int>& nums, int start, int end) {
        if (end - start == 0) return nums[start];
        int pre = 0;
        int cur = nums[start];
        for (int i = start + 1; i < end + 1; ++i) {
            int t = max(pre + nums[i], cur);
            pre = cur;
            cur = t;
        }
        return cur;
    }
};

Go

func rob(nums []int) int {
    n := len(nums)
    if n == 0 {
        return 0
    }
    return robRange(nums, 0, n - 1)
}

func robRange(nums[]int, start int, end int) int {
    if end - start == 0 {
        return nums[start]
    }
    pre, cur := 0, nums[start]
    for i := start + 1; i < end + 1; i++ {
        pre, cur = cur, max(pre + nums[i], cur)
    }
    return cur
}

func max(a, b int) int {
    if (a > b) {
        return a
    }
    return b
}

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