LeetCode:House Robber
Total Accepted: 41584
Total Submissions: 132317
Difficulty: Easy
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.
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思路：
1.假设nums=[ 3 2 6 5 6 3]
dp[0]=3,
dp[1]=3;
dp[2]=max(nums[1],nums[0]+nums[2]);
...
dp[i]=max(dp[i-2]+nums[i], dp[i-1])
2.得到 dp=[3 3 9 9 15 15]
code:
class Solution {
public:
int rob(vector<int>& nums) {
int len = nums.size();
if(0==len) return 0;
int dp[len];
if(len>0) dp[0] = nums[0];
if(len>1) dp[1] = max(nums[0],nums[1]);
for(int i=2;i<len;i++)
else dp[i]=max(dp[i-2]+nums[i],dp[i-1]);
return dp[len-1];
}
};
空间优化后：
class Solution {
public:
int rob(vector<int>& nums) {
int len = nums.size();
if(0==len) return 0;
int preN=0;
int preY=0;
for(int i=1;i<len;i++){
int tmp = preY;
preN = max(preN, preY);
preY = tmp + nums[i];
}
return max(preN, preY);
}
};