整体思路就是二分法查找,在处理mid时比较下mid与mid+1的关系,看是否是递增还是递减。 java代码如下: 1publicclassSolution {2publicintfindPeakElement(int[] num) {3intstart=0,end=num.length-1,mid=0,mid1=0;4while(start<end){5mid=(start+end)/2;6mid1=mid+1;7if(num[mid]<num[mid1]) start...
Your solution should be in logarithmic complexity. 思路: 可能包含多个极值点(peak),只要取任意一个就行。假设只有一个peak,用二分查找根据mid元素单增单减的关系可以判断peak在mid的左边还是右边。如果不止一个peak,用二分法划分也不会将查找区间[left,right]划到没有peak的区间。时间复杂度为O(logn)。 算法:...
classSolution {publicintfindPeakElement(int[] nums) {if(nums.length==0 || nums==null)return0;intL = 0;intR = nums.length - 1;while(L<R){intmid = (L+R+1)>>>1;if(nums[mid-1]>nums[mid]){ R= mid-1; }else{ L=mid; } }returnL; } }...
直接想到的思路是,由于num[-1]是负无穷,因此num[0]对于它的previous neighbor是上升的,只需从num头开始遍历,找到第一个num[i] > num[i + 1]的数,那么i就是Peak Element; 算法虽然看上去已经很高效,但是,时间复杂度o(n); 代码: 1classSolution {2public:3intfindPeakElement(constvector<int> &num) {4...
【leetcode】Find Peak Element Find Peak Element 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 ...
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<...
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.
Explanation: Your function can return either index number 1 where the peak element is 2, or index number 5 where the peak element is 6. Note: Your solution should be in logarithmic complexity. 解题思路:如果当前元素小于前一个元素,则前一个元素应当为峰值,否则继续向后遍历。
For example, in array[1, 2, 3, 1], 3 is a peak element and your function should return the index number 2. 二分查找! 1classSolution {2public:3intfindPeak(constvector<int> &num,intleft,intright) {4if(left == right)returnleft;5intmid1 = (left + right) /2;6intmid2 = mid1...
https://leetcode.com/problems/find-peak-element/二.题目大意:给定一个长度为N的一维数组,数组是无序的,要求找到数组中的极大值(或局部最大值),并返回该极大值的下标,并假设 nums[-1] = nums[n] = -∞.;当某元素同时大于它两边的元素时,则该元素是数组中的一个极大值,数组中若存在多个极大值,则...