From b16bf4499a1e359860771568fe3f07bccd0cd80b Mon Sep 17 00:00:00 2001
From: yujiahan2018 <yujiahan_biancheng@163.com>
Date: Sat, 20 May 2023 19:31:51 +0800
Subject: [PATCH] =?UTF-8?q?Update=200198.=E6=89=93=E5=AE=B6=E5=8A=AB?=
 =?UTF-8?q?=E8=88=8D.md?=
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---
 problems/0198.打家劫舍.md | 35 ++++++++++++++++++++++++++++++++---
 1 file changed, 32 insertions(+), 3 deletions(-)

diff --git a/problems/0198.打家劫舍.md b/problems/0198.打家劫舍.md
index 6e682ec3..7c8ba4db 100644
--- a/problems/0198.打家劫舍.md
+++ b/problems/0198.打家劫舍.md
@@ -141,7 +141,36 @@ class Solution {
 	}
 }
 
-// 空间优化 dp数组只存与计算相关的两次数据
+// 使用滚动数组思想，优化空间
+// 分析本题可以发现，所求结果仅依赖于前两种状态，此时可以使用滚动数组思想将空间复杂度降低为3个空间
+class Solution {
+    public int rob(int[] nums) {
+        
+        int len = nums.length;
+
+        if (len == 0) return 0;
+        else if (len == 1) return nums[0];
+        else if (len == 2) return Math.max(nums[0],nums[1]);
+
+
+        int[] result = new int[3]; //存放选择的结果
+        result[0] = nums[0];
+        result[1] = Math.max(nums[0],nums[1]);
+        
+
+        for(int i=2;i<len;i++){
+
+            result[2] = Math.max(result[0]+nums[i],result[1]);
+
+            result[0] = result[1];
+            result[1] = result[2];
+        }
+        
+        return result[2];
+    }
+}
+
+// 进一步对滚动数组的空间优化 dp数组只存与计算相关的两次数据
 class Solution {
     public int rob(int[] nums) {
         if (nums.length == 1)  {
@@ -151,11 +180,11 @@ class Solution {
         // 优化空间 dp数组只用2格空间 只记录与当前计算相关的前两个结果
         int[] dp = new int[2];
         dp[0] = nums[0];
-        dp[1] = nums[0] > nums[1] ? nums[0] : nums[1];
+        dp[1] = Math.max(nums[0],nums[1]);
         int res = 0;
         // 遍历
         for (int i = 2; i < nums.length; i++) {
-            res = (dp[0] + nums[i]) > dp[1] ? (dp[0] + nums[i]) : dp[1];
+            res = Math.max((dp[0] + nums[i]) , dp[1] );
             dp[0] = dp[1];
             dp[1] = res;
         }