Can you solve this real interview question? Find Peak Element - A peak element is an element that is strictly greater than its neighbors. Given a 0-indexed integer array nums, find a peak element, and return its index. If the array contains multiple pea
A peak element is an element that is greater than its neighbors. num[i] ≠ num[i+1], find a peak element and return its index. The array may contain multiple peaks, in that case return the index to any one of the peaks is fine. num[-1] = num[n] = -∞.[1, 2, 3, 1], ...
思路很简单:在Binary Search上略加修改。规则是:只要mid比其左邻居小,那说明左半段必然存在peak,只访问左半段即可;否则,右半段必然存在peak。需要注意的是,两端也可以是peak。 1classSolution {2public:3intfindPeakElement(constvector<int> &num) {4intn=num.size();5intleft=0,right=n-1;6while(left<...
leetcode -- Find Peak Element -- 找波峰--重点 https://leetcode.com/problems/find-peak-element/ 思路就是二分查找。如果mid在波峰的递减部分,那么e = mid - 1. 如果mid在波峰的递增部分,那么s = mid + 1. 如果mid在波峰,那么返回。 my code: 效率不高。 class Solution(object): def findPeakE...
A peak element is an element that is greater than its neighbors. Given an input array wherenum[i] ≠ num[i+1], find a peak element and return its index. The array may contain multiple peaks, in that case return the index to any one of the peaks is fine. ...
leetcode-34. Find First and Last Position of Element in Sorte-binary-searchyjhycl 立即播放 打开App,流畅又高清100+个相关视频 更多 84 0 18:47 App leetcode-222-Counter Complete Binary Tree Nodes - binary search 2 0 10:00 App leetcode-852. Peak Index in a Mountain Array -binary-search...
http://siddontang.gitbooks.io/leetcode-solution/content/array/find_peak_element.html Paste_Image.png ** 总结: Array 凡是涉及到搜索的,而且要求复杂度是 log n 的, 一般都会涉及到, Binary Search!!! ** Anyway, Good luck, Richardo! 写出...
(midVal>leftVal&&midVal>rightVal){midIndex}elseif(midVal<leftVal){findPeakElement(nums,startIndex,leftIndex)}else{findPeakElement(nums,rightIndex,endIndex)}}}funmain(args:Array<String>){valinput=intArrayOf(1,2,4,2)valfindPeakElement=FindPeakElement()println(findPeakElement.findPeakElement(...
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这种找Peak Element的首先想到的是binary search, 因为有更优的时间复杂度,否则brute forceO(n)的方法太直接了。 binary search的方法就是比较当前element与邻居,至于左邻居还是右邻居都可以,只要一致就行。不断缩小范围,最后锁定peak element. 这种类似题也有很多Follow up, 比如先增后减的数组怎么找指定元素,甚至...