162. 寻找峰值
作者:互联网
package leetcode; //nums[mid+1]>nums[mid],那么mid+2只有两种可能,一个是大于mid+1,一个是小于mid+1,小于mid+1的情况,那么mid+1就是峰值, //大于mid+1的情况,继续向右推,如果一直到数组的末尾都是大于的,那么可以肯定最后一个元素是峰值,因为nums[nums.length]=负无穷 //num[mid+1]<nums[mid]同理 public class demo_162 { public int findPeakElement(int[] nums) { int left=0; int right=nums.length-1; int mid=0; while(left<=right) { mid=(left+right)/2; //nums[mid+1]>nums[mid]情况 if((mid+1)<nums.length&&nums[mid]<nums[mid+1]) { if((mid+2)==nums.length||nums[mid+2]<nums[mid+1]) { return (mid+1); } else { left=mid+1; } } //nums[mid+1]<nums[mid]情况 else { if((mid-1)>-1&&nums[mid]>nums[mid-1]) { return mid; } else { right=mid-1; } } } return mid; } public static void main(String[] args) { // TODO Auto-generated method stub demo_162 d162=new demo_162(); int[] nums= {1,2}; System.out.println(d162.findPeakElement(nums)); } }
标签:nums,demo,mid,峰值,寻找,d162,162 来源: https://www.cnblogs.com/Yshun/p/15087021.html