[pull] master from wisdompeak:master #354

Original file line number	Diff line number	Diff line change
Expand Up		@@ -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)
Expand Down

23 changes: 23 additions & 0 deletions Thinking/3660.Jump-Game-IX/3660.Jump-Game-IX.cpp

Show comments View file Open in desktop

Original file line number	Diff line number	Diff line change
		@@ -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;
	}
	};

11 changes: 11 additions & 0 deletions Thinking/3660.Jump-Game-IX/Readme.md

Show comments View file Open in desktop

Original file line number	Diff line number	Diff line change
		@@ -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。

Navigation Menu