ID | Title | Difficulty | |
---|---|---|---|
Loading... |
198. House Robber
Medium
LeetCode
Array, Dynamic Programming
Problem
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.
Code
class Solution {
public int rob(int[] nums) {
if(nums == null || nums.length == 0) return 0;
int[] dp = new int[nums.length];
// 要处理两个特殊情况
dp[0] = nums[0];
if(nums.length == 1) return dp[0];
dp[1] = Math.max(nums[0], nums[1]);
if(nums.length == 2) return dp[1];
for(int i = 2; i < nums.length; i++){
dp[i] = Math.max(nums[i] + dp[i-2], dp[i-1]);
}
return dp[nums.length - 1];
}
}
public int rob(int[] num) {
int prevMax = 0;
int currMax = 0;
for (int x : num) {
int temp = currMax;
currMax = Math.max(prevMax + x, currMax);
prevMax = temp;
}
return currMax;
}
按 <- 键看上一题!
197. Rising Temperature
按 -> 键看下一题!
199. Binary Tree Right Side View