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 <= 1000 <= nums[i] <= 400
解法关键点： 截止当前房子能偷的最大值是由前面两间房子决定的, 如果偷curr, curr-1就不能偷，那么结果就是 mem[curr]+mem[curr-2] ； 如果不偷curr，那么结果就是mem[curr-1]。我们需要取二者最大值，决定是否偷当前房子
mem[i]=Math.max(mem[i-1],nums[i]+mem[i-2]);
class Solution {
public int rob(int[] nums) {
if(nums.length==1) return nums[0];
if(nums.length==2) return Math.max(nums[0],nums[1]);
int[] mem = new int[nums.length];
mem[0]=nums[0];
mem[1]=Math.max(nums[0],nums[1]);
for(int i=2;i<nums.length;i++){
mem[i]=Math.max(mem[i-1],nums[i]+mem[i-2]);
}
return mem[nums.length-1];
}
}
213. House Robber II
Medium
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 a security system 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 = [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: 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 3:
Input: nums = [1,2,3] Output: 3
Constraints:
1 <= nums.length <= 1000 <= nums[i] <= 1000
解法:这题就是前一道的套娃题，分两种情况偷：1.第一间不偷 2.最后一间不偷
class Solution {
public int rob(int[] nums) {
if(nums.length==1) return nums[0];
if(nums.length==2) return Math.max(nums[0],nums[1]);
return Math.max(rob(nums,0,nums.length-2),rob(nums,1,nums.length-1));
}
private int rob(int[] nums,int start,int end){
int[] mem = new int[nums.length];
mem[start]=nums[start];
mem[start+1]=Math.max(nums[start],nums[start+1]);
for(int i=start+2;i<=end;i++){
mem[i]=Math.max(mem[i-2]+nums[i],mem[i-1]);
}
return mem[end];
}
}