首页 > 代码库 > Find Peak Element
Find Peak Element
Find Peak Element
A peak element is an element that is greater than its neighbors.
Given an input array where 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.
You may imagine that num[-1] = num[n] = -∞
.
For example, in array [1, 2, 3, 1]
, 3 is a peak element and your function should return the index number 2.
我自己写的就比较挫一点了,这道题要用二分查找来实现。但我一直没有明白为什么可以用二分查找实现
1 public class Solution { 2 public int findPeakElement(int[] num) { 3 if(num.length == 0 || num.length == 1) 4 return 0; 5 boolean isIncr = false; //是否为升序 6 int index = -1; //peak索引值 7 int numArray[] = new int[num.length + 2]; 8 for(int i = 0; i < num.length;i++){ 9 numArray[i + 1] = num[i];10 }11 numArray[0] = Integer.MIN_VALUE;12 numArray[numArray.length - 1] = Integer.MIN_VALUE;13 14 for(int i = 1; i < numArray.length;i++){15 if(numArray[i] > numArray[i - 1])16 isIncr = true;17 if(isIncr && numArray[i] < numArray[i - 1]){18 index = i - 2;19 break;20 }21 }22 23 return index;24 }25 }
下面这个是百度到的,好像是CSDN还是哪儿的,我就懒得去找源出处了。这个时间复杂度是O(N)
1 public class Solution {2 public int findPeakElement(int[] num) {3 for(int i = 1; i < num.length; i++){4 if(num[i] < num[i - 1])5 return i - 1;6 }7 return num.length - 1;8 }9 }
下面这个是讨论区用二分查找实现的
1 public class Solution { 2 public int findPeakElement(int[] num) { 3 int low = 0; 4 int high = num.length - 1; 5 while (low < high) { 6 int mid = (low + high + 1)/2; 7 if (mid == 0 || num[mid] > num[mid-1]) { 8 low = mid; 9 } else {10 high = mid-1;11 }12 }13 return low;14 }15 }
Find Peak Element
声明:以上内容来自用户投稿及互联网公开渠道收集整理发布,本网站不拥有所有权,未作人工编辑处理,也不承担相关法律责任,若内容有误或涉及侵权可进行投诉: 投诉/举报 工作人员会在5个工作日内联系你,一经查实,本站将立刻删除涉嫌侵权内容。