[leetcode]-213. House Robber II（C语言）

最新推荐文章于 2020-03-31 22:43:06 发布

原创最新推荐文章于 2020-03-31 22:43:06 发布 · 254 阅读

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本文探讨了一种特殊的抢劫问题：假设房子排列成环形，且相邻的房子有相连的安全系统，如何在不触动警报的情况下获得最大金额？通过动态规划算法解决此问题。

You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed. All houses at this place are arranged in a circle. That means the first house is the neighbor of the last one. Meanwhile, 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: [2,3,2]
Output: 3
Explanation: You cannot rob house 1 (money = 2) and then rob house 3 (money = 2),
             because they are adjacent houses.

Example 2:

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.

int rob(int* nums, int numsSize) {
    if(numsSize<=0)
        return 0;
    if(numsSize==1)
        return nums[0];
    if(numsSize==2)
        return nums[0]>nums[1]?nums[0]:nums[1];
    int n1,n2,i,j,k,dp1[numsSize],dp2[numsSize];
    dp1[0]=nums[0];
    dp1[1]=nums[1]>nums[0]?nums[1]:nums[0];
    for(i=2;i<numsSize-1;i++)
    {
        dp1[i]=nums[i]+dp1[i-2]>dp1[i-1]?nums[i]+dp1[i-2]:dp1[i-1];
    }
    n1=dp1[i-1];
    dp2[1]=nums[1];
    dp2[2]=nums[1]>nums[2]?nums[1]:nums[2];
    for(i=3;i<numsSize;i++)
        dp2[i]=nums[i]+dp2[i-2]>dp2[i-1]?nums[i]+dp2[i-2]:dp2[i-1];
    n2=dp2[i-1];
    return n1>n2?n1:n2;
}