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.
- 动态规划，设置maxV[i]表示到第i个房子位置，最大收益。
- 递推关系为maxV[i] = max(maxV[i-2]+num[i], maxV[i-1])
class Solution {
public:
int rob(vector<int> &num) {
int n = num.size();
if(n == )
return ;
else if(n == )
return num[];
else
{
vector<int> maxV(n, );
maxV[] = num[];
maxV[] = max(num[], num[]);
for(int i = ; i < n; i ++)
maxV[i] = max(maxV[i-]+num[i], maxV[i-]);
return maxV[n-];
}
}
};
- 用A[0]表示没有rob当前house的最大money，A[1]表示rob了当前house的最大money，那么A[0] 等于rob或者没有rob上一次house的最大值
- 即A[i+1][0] = max(A[i][0], A[i][1]).. 那么rob当前的house，只能等于上次没有rob的+money[i+1], 则A[i+1][1] = A[i][0]+money[i+1]. 只需要两个变量保存结果就可以了
int max(int a, int b)
{
if(a > b)
return a;
else
return b;
}
int rob(int* nums, int numsSize) {
int best0 = ; // 表示没有选择当前houses
int best1 = ; // 表示选择了当前houses
for(int i = ; i < numsSize; i++){
int temp = best0;
best0 = max(best0, best1); // 没有选择当前houses，那么它等于上次选择了或没选择的最大值
best1 = temp + nums[i]; // 选择了当前houses，值只能等于上次没选择的+当前houses的money
}
return max(best0, best1);
}
After robbing those houses on that street, the thief has found himself a new place for his thievery so that he will not get too much attention. This time, all houses at this place are arranged in a circle. That means the first house is the neighbor of the last one. Meanwhile, the security system for these houses remain the same as for those in the previous street.
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.
- 把第一个房子，作为例外处理
- 如果选择第一个房子，则从第三个房子开始到倒数第二房子，属于常规
- 如果不选第一个房子，则从第二个房子开始属于常规
int rob(int* nums, int numsSize) {
if(numsSize == )
return ;
//分成第一家抢和不抢
//如果第一家不抢，最后一个就不用考虑抢不抢
int hos11 = nums[];
int hos10 = ;
int best1 = ; //rub cur house
int best0 = ; //not rub cur house
for(int i = ; i < numsSize - ; i++){
int tmp = best0;
best0 = best1 > best0 ? best1 : best0;
best1 = tmp + nums[i];
}
hos11 += best1 > best0 ? best1 : best0;
//如果第一家抢，最后一家不能抢，返回倒数第二家的情况
best1 = ;
best0 = ;
for(int i = ; i < numsSize; i++){
int tmp = best0;
best0 = best1 > best0 ? best1 : best0;
best1 = tmp + nums[i];
}
hos10 = best1 > best0 ? best1 : best0;
return hos11 > hos10 ? hos11 : hos10;
}