From e9f7683d4679a6126907134ba4c623267b1ba951 Mon Sep 17 00:00:00 2001
From: LiHua <1985390347@qq.com>
Date: Wed, 24 Nov 2021 10:11:18 +0800
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---
 .../0188.买卖股票的最佳时机IV.md     | 33 +++++++++++++------
 1 file changed, 23 insertions(+), 10 deletions(-)

diff --git a/problems/0188.买卖股票的最佳时机IV.md b/problems/0188.买卖股票的最佳时机IV.md
index 73861b35..0e898b43 100644
--- a/problems/0188.买卖股票的最佳时机IV.md
+++ b/problems/0188.买卖股票的最佳时机IV.md
@@ -222,19 +222,32 @@ class Solution {
 //版本三：一维 dp数组
 class Solution {
     public int maxProfit(int k, int[] prices) {
-        //在版本二的基础上，由于我们只关心前一天的股票买入情况，所以只存储前一天的股票买入情况
-        if(prices.length==0)return 0;
-        int[] dp=new int[2*k+1];
-        for (int i = 1; i <2*k ; i+=2) {
-            dp[i]=-prices[0];
+        if(prices.length == 0){
+            return 0;
         }
-        for (int i = 0; i <prices.length ; i++) {
-            for (int j = 1; j <2*k ; j+=2) {
-                dp[j]=Math.max(dp[j],dp[j-1]-prices[i]);
-                dp[j+1]=Math.max(dp[j+1],dp[j]+prices[i]);
+        if(k == 0){
+            return 0;
+        }
+        // 其实就是123题的扩展，123题只用记录2天的状态
+        // 这里记录k天的状态就行了
+        // 每天都有买入，卖出两个状态，所以要乘 2
+        int[] dp = new int[2 * k];
+        // 按123题解题格式那样，做一个初始化
+        for(int i = 0; i < dp.length / 2; i++){
+            dp[i * 2] = -prices[0];
+        }
+        for(int i = 1; i <= prices.length; i++){
+            dp[0] = Math.max(dp[0], -prices[i - 1]);
+            dp[1] = Math.max(dp[1], dp[0] + prices[i - 1]);
+            // 还是与123题一样，与123题对照来看
+            // 就很容易啦
+            for(int j = 2; j < dp.length; j += 2){
+                dp[j] = Math.max(dp[j], dp[j - 1] - prices[i-1]);
+                dp[j + 1] = Math.max(dp[j + 1], dp[j] + prices[i - 1]);
             }
         }
-        return dp[2*k];
+        // 返回最后一天卖出状态的结果就行了
+        return dp[dp.length - 1];
     }
 }
 ```