198. House Robber(打家劫舍)(求不相邻的位置上的数字之和的最大值)
198. House Robber(打家劫舍)(求不相邻的位置上的数字之和的最大值)
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.你是一个专业的小偷，计划偷窃沿街的房屋。每间房内都藏有一定的现金，影响你偷窃的唯一制约因素就是相邻的房屋装有相互连通的防盗系统，如果两间相邻的房屋在同一晚上被小偷闯入，系统会自动报警。
给定一个代表每个房屋存放金额的非负整数数组，计算你在不触动警报装置的情况下，能够偷窃到的最高金额。
示例 1:
输入: [1,2,3,1]
输出: 4
解释: 偷窃 1 号房屋 (金额 = 1) ，然后偷窃 3 号房屋 (金额 = 3)。
偷窃到的最高金额 = 1 + 3 = 4 。示例 2:
输入: [2,7,9,3,1]
输出: 12
解释: 偷窃 1 号房屋 (金额 = 2), 偷窃 3 号房屋 (金额 = 9)，接着偷窃 5 号房屋 (金额 = 1)。
偷窃到的最高金额 = 2 + 9 + 1 = 12 。class Solution {
public int rob(int[] nums) {
if (nums == null || nums.length == 0) return 0;
int len = nums.length;
if (len == 1) return nums[0];
if (len == 2) return Math.max(nums[0], nums[1]);
int[] dp = new int[len];
dp[0] = nums[0];
dp[1] = Math.max(dp[0], nums[1]); // 在dp[1]这里能够积累的最大价值
for (int i = 2; i < len; ++i) {
dp[i] = Math.max(dp[i - 1], dp[i - 2] + nums[i]); // 下标为i的dp积累的最大价值
}
return dp[len - 1];
}
}
题目用例有[2, 1, 1, 2]
转化成原型就是：给定一个正数数组，求不相邻的位置上的数字之和的最大值。(可能间隔一个间隔多个)
Debug code in playground:
class Solution {
public int rob(int[] nums) {
if (nums == null || nums.length == 0) return 0;
int len = nums.length;
if (len == 1) return nums[0];
if (len == 2) return Math.max(nums[0], nums[1]);
int[] dp = new int[len];
dp[0] = nums[0];
dp[1] = Math.max(dp[0], nums[1]); // 在dp[1]这里能够积累的最大价值
for (int i = 2; i < len; ++i) {
dp[i] = Math.max(dp[i - 1], dp[i - 2] + nums[i]); // 下标为i的dp积累的最大价值
}
return dp[len - 1];
}
}
public class MainClass {
public static int[] stringToIntegerArray(String input) {
input = input.trim();
input = input.substring(1, input.length() - 1);
if (input.length() == 0) {
return new int[0];
}
String[] parts = input.split(",");
int[] output = new int[parts.length];
for(int index = 0; index < parts.length; index++) {
String part = parts[index].trim();
output[index] = Integer.parseInt(part);
}
return output;
}
public static void main(String[] args) throws IOException {
BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
String line;
while ((line = in.readLine()) != null) {
int[] nums = stringToIntegerArray(line);
int ret = new Solution().rob(nums);
String out = String.valueOf(ret);
System.out.print(out);
}腾讯云开发者
