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567jin
AlgorithmPractice
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_198_Rob.java
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48 lines (45 loc) · 1.73 KB
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_198_Rob.java
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/**
* @author jzy
* @GitHub <a href="https://github.com/567jin">567jin</a>
* @since 2024-03-14 10:39:04
*/
public class _198_Rob {
public static void main(String[] args) {
int[] nums = {1, 2, 3, 1};
int rob = new _198_Rob().rob(nums);
System.out.println(rob);
int rob2 = new _198_Rob().rob2(nums);
System.out.println(rob2);
}
/**
* dp[i]的定义是从i个房子中 偷盗的最大金额
* dp[i] = max(dp[i-1], dp[i-2])+nums[i] 推导错误 分析是正确的就是连续的两间房子选一个 但是没有将公式推导明白
* 正确的公式是 dp[i]=max(dp[i-1](偷了i-1家那么第i家就不能偷了),dp[i-2]+nums[i-1](偷前i-2间加上当前这个房子的钱,注意nums中的i从0开始 dp中的i从1开始)) dp[0]=0表示第0家 dp[1]表示偷盗第一家
* <p>
* 每次只能选择两个房屋中的一个进行偷盗
*/
public int rob(int[] nums) {
int n = nums.length;
int[] dp = new int[n + 1];
// 初始化
dp[0] = 0; // i为0 表示没有房子 初始化为0
dp[1] = nums[0]; // i为1 只有一间房子 只能偷这一间
for (int i = 2; i <= n; i++) {
dp[i] = Math.max(dp[i - 1], dp[i - 2] + nums[i - 1]); // 要么偷当前的(上一件没偷过dp[i-2]) 要么不偷(上一间偷过了dp[i-1])
}
return dp[n];
}
/**
* 空间优化
*/
public int rob2(int[] nums) {
int maxMoney = 0; // dp[i-1]
int prev = 0; // dp[i-2]
for (int num : nums) {
int temp = maxMoney; // dp[i-2]的结果
maxMoney = Math.max(prev + num, maxMoney);
prev = temp;
}
return maxMoney;
}
}
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