diff --git a/Readme.md b/Readme.md
index 70938ab70..53da52feb 100644
--- a/Readme.md
+++ b/Readme.md
@@ -1704,6 +1704,7 @@
 [2952.Minimum-Number-of-Coins-to-be-Added](https://github.com/wisdompeak/LeetCode/tree/master/Thinking/2952.Minimum-Number-of-Coins-to-be-Added) (H-) 
 [3609.Minimum-Moves-to-Reach-Target-in-Grid](https://github.com/wisdompeak/LeetCode/tree/master/Thinking/3609.Minimum-Moves-to-Reach-Target-in-Grid) (H) 
 [3644.Maximum-K-to-Sort-a-Permutation](https://github.com/wisdompeak/LeetCode/tree/master/Thinking/3644.Maximum-K-to-Sort-a-Permutation) (H) 
+[3660.Jump-Game-IX](https://github.com/wisdompeak/LeetCode/tree/master/Thinking/3660.Jump-Game-IX) (H) 
 
 #### [LeetCode Cup](https://github.com/wisdompeak/LeetCode/tree/master/LCCUP)
 [LCP23.魔术排列](https://github.com/wisdompeak/LeetCode/tree/master/LCCUP/2020Fall/LCP23.%E9%AD%94%E6%9C%AF%E6%8E%92%E5%88%97) 
diff --git a/Thinking/3660.Jump-Game-IX/3660.Jump-Game-IX.cpp b/Thinking/3660.Jump-Game-IX/3660.Jump-Game-IX.cpp
new file mode 100644
index 000000000..ce95f215a
--- /dev/null
+++ b/Thinking/3660.Jump-Game-IX/3660.Jump-Game-IX.cpp
@@ -0,0 +1,23 @@
+class Solution {
+public:
+ vector<int> maxValue(vector<int>& nums) {
+ int n = nums.size();
+ vector<int>preMax(n);
+ vector<int>sufMin(n);
+ for (int i=0; i<n; i++) + preMax[i] = max(i==0?0:preMax[i-1], nums[i]); + for (int i=n-1; i>=0; i--)
+ sufMin[i] = min((i==n-1)?INT_MAX:sufMin[i+1], nums[i]);
+
+ vector<int>rets(n);
+ rets[n-1] = preMax[n-1];
+ for (int i=n-2; i>=0; i--) {
+ if (preMax[i]>sufMin[i+1])
+ rets[i] = rets[i+1];
+ else
+ rets[i] = preMax[i];
+ } 
+
+ return rets;
+ }
+};
diff --git a/Thinking/3660.Jump-Game-IX/Readme.md b/Thinking/3660.Jump-Game-IX/Readme.md
new file mode 100644
index 000000000..31b1a3df6
--- /dev/null
+++ b/Thinking/3660.Jump-Game-IX/Readme.md
@@ -0,0 +1,11 @@
+### 3660.Jump-Game-IX
+
+此题乍看是个图论的问题,彼此可以跳转的点可以认为是联通的。但是构建所有的边需要n^2的复杂度。
+
+我们定义preMax[i]表示前i个元素(包括自身)里的最大值。考察任意的rets[i],如果答案只在左边的话,那么答案就是preMax[i]。但是也有可能答案在右边。从i能往右跳转到哪些地方呢?我们势必会先借助于preMax[i],因为从preMax[i]跳转的话可以最大范围地覆盖到[i+1:n-1]里的可跳转区域。此时两种情况:
+
+1. 如果`preMax[i]<sufmin[i+1]`,也就是说从premax[i]也无法跳转到i右边的任何位置,于是rets[i]只能是premax[i]. + +2. 如果`preMax[i]>sufMin[i+1]`,那么我们可以有这样一条跳转路径:i->preMax[i]->sufMin[i]->i+1. 最后一步跳转的依据是:根据定义,sufMin[i]是[i+1:n-1]里的最小值,必然小于等于nums[i+1].由此可知i与i+1是联通的,必然有`rets[i]=rets[i+1]`.
+
+由此我们发现了rets从后往前的递推关系。因为rets[n-1]必然等于preMax[n-1],由此可以根据以上的结论,依次推出i=n-2,n-1,...,0的答案。
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