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Simplify House Robber solution

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Yutong Wang 2018-09-17 21:50:19 -07:00 committed by GitHub
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notes/Leetcode 题解.md
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@ -2415,27 +2415,19 @@ public int climbStairs(int n) {
定义 dp 数组用来存储最大的抢劫量，其中 dp[i] 表示抢到第 i 个住户时的最大抢劫量。 定义 dp 数组用来存储最大的抢劫量，其中 dp[i] 表示抢到第 i 个住户时的最大抢劫量。
由于不能抢劫邻近住户，因此如果抢劫了第 i 个住户那么只能抢劫 i - 2 或者 i - 3 的住户，所以 由于不能抢劫邻近住户，因此如果抢劫了第 i 个住户那么只能抢劫 i - 2 或者 i - 3 的住户，所以
dp[i] = max(dp[i-1], dp[i-2] + nums[i]) <br>
<div align="center"><img src="https://latex.codecogs.com/gif.latex?dp[i]=max(dp[i-2],dp[i-3])+nums[i]"/></div> <br>
```java ```java
public int rob(int[] nums) { public int rob(int[] nums) {
int n = nums.length; int pre2 = 0, pre1 = 0;
if (n == 0) { for (int i = 0; i < nums.length; i++) {
return 0; int cur = Math.max(pre2 + nums[i], pre1);
}
if (n == 1) {
return nums[0];
}
int pre3 = 0, pre2 = 0, pre1 = 0;
for (int i = 0; i < n; i++) {
int cur = Math.max(pre2, pre3) + nums[i];
pre3 = pre2;
pre2 = pre1; pre2 = pre1;
pre1 = cur; pre1 = cur;
} }
return Math.max(pre1, pre2); return pre1;
} }
``` ```
**强盗在环形街区抢劫** **强盗在环形街区抢劫**
@ -2443,7 +2435,7 @@ public int rob(int[] nums) {
[213. House Robber II (Medium)](https://leetcode.com/problems/house-robber-ii/description/) [213. House Robber II (Medium)](https://leetcode.com/problems/house-robber-ii/description/)
```java ```java
public int rob(int[] nums) { public int rob(int[] nums) {
if (nums == null || nums.length == 0) { if (nums == null || nums.length == 0) {
return 0; return 0;
} }
@ -2454,15 +2446,14 @@ public int rob(int[] nums) {
return Math.max(rob(nums, 0, n - 2), rob(nums, 1, n - 1)); return Math.max(rob(nums, 0, n - 2), rob(nums, 1, n - 1));
} }
private int rob(int[] nums, int first, int last) { private int rob(int[] nums, int first, int last) {
int pre3 = 0, pre2 = 0, pre1 = 0; int pre2 = 0, pre1 = 0;
for (int i = first; i <= last; i++) { for (int i = first; i <= last; i++) {
int cur = Math.max(pre3, pre2) + nums[i]; int cur = Math.max(pre1, pre2 + nums[i]);
pre3 = pre2;
pre2 = pre1; pre2 = pre1;
pre1 = cur; pre1 = cur;
} }
return Math.max(pre2, pre1); return pre1;
} }
``` ```