[leetcode]-198. House Robber（C语言）

最新推荐文章于 2022-04-10 15:21:15 发布

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本文探讨了一种算法问题——专业窃贼计划沿街抢劫房屋，但相邻房屋间的安全系统会互相联动报警。通过给定每户人家藏匿的金钱数额列表，目标是找出最大抢劫金额而不触发报警的方法。文章提供了具体的例子并附带了解决方案的C语言实现。

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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.

Example 1:

Input: [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: [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.

int rob(int* nums, int numsSize) {
    if(numsSize<=0)
        return 0;
    if(numsSize==1)
        return nums[0];
    int dp[numsSize],j;
    dp[0]=nums[0];
    dp[1]=(nums[0]>=nums[1]?nums[0]:nums[1]);
    for(j=2;j<numsSize;j++)
    {
        dp[j]=(nums[j]+dp[j-2]>dp[j-1]?nums[j]+dp[j-2]:dp[j-1]);
    }
    return dp[j-1];
}