Leetcode之二分法专题-852. 山脉数组的峰顶索引(Peak Index in a Mountain Array) 我们把符合下列属性的数组A称作山脉: A.length >= 3 存在0 < i < A.length - 1使得A[0] < A[1] < ... A[i-1] < A[i] > A[i+1] > ... > A[A.length - 1] 给定一个确定为山脉的数组,返回任何满足...
A is a mountain, as defined above.二分法:class Solution{ public: int peakIndexInMountainArray(vector<int>& a){ int beg = 1,end = a.size(); int mid = (beg + end) / 2; while(beg <= end){ if(a[mid] < a[mid - 1]){ end = mid - 1; } else if(a[mid] < a[mid + ...
代码实现 class Solution: def peakIndexInMountainArray(self, A): """ :type A: List[int] :rtype: int """ return A.index(max(A)) 1. 2. 3. 4. 5. 6. 7.
LeetCode 852. Peak Index in a Mountain Array 2019-12-21 09:43 −原题链接在这里:https://leetcode.com/problems/peak-index-in-a-mountain-array/ 题目: Let's call an array A a mountain if the following proper... Dylan_Java_NYC
https://github.com/grandyang/leetcode/issues/852 类似题目: Find Peak Element 参考资料: https://leetcode.com/problems/peak-index-in-a-mountain-array/ https://leetcode.com/problems/peak-index-in-a-mountain-array/discuss/139848/C%2B%2BJavaPython-Better-than-Binary-Search ...
"852. Peak Index in a Mountain Array Easy" 方法一:二分查找 官方答案: java class Solution { public int peakIndexInMountainArray(int[] A) { int lo = 0, h
今天介绍的是LeetCode算法题中Easy级别的第199题(顺位题号是852)。如果以下属性成立,我们将数组A称为山: A.length> = 3。 存在一个i(0 < i < A.length-1),使得A[0] <A[1] <... A[i-1] < A[i] > A[i + 1]> ...> A[A.length - 1]。
LeetCode 852 Peak Index in a Mountain Array 解题报告 题目要求 Let's call an arrayAa mountain if the following properties hold: A.length >= 3 There exists some0 < i < A.length - 1such thatA[0] < A[1] < ... A[i-1] < A[i] > A[i+1] > ... > A[A.length - 1]...
leetcode.852.PeakIndexinaMountainArray links: https://leetcode.com/problems/peak-index-in-a-mountain-array/ 我的思路,直接遍历查找,找到第一个变小的数的位置,之前的那个位置就目的index 1 2 3 4 5 6 7 8 9 10 11 12 classSolution(object):...
LeetCode 852. Peak Index in a Mountain Array 原题链接在这里:https://leetcode.com/problems/peak-index-in-a-mountain-array/ 题目: Let's call an arrayAamountainif the following properties hold: A.length >= 3 There exists some0 < i < A.length - 1such thatA[0] < A[1] < ... A...