For example, in array[1, 2, 3, 1], 3 is a peak element and your function should return the index number 2. 链接:http://leetcode.com/problems/find-peak-element/ 题解: 找1D-Peak,正好是MIT 6.006第一课的第一个问题。使用Binary Search,当nums[mid] < nums[mid + 1],nums[mid + 1]...
binary search寻找peak element,因为初始数组的开头是升序,结尾是降序,所以一定存在peak,每一次搜索如果中间点mid仍旧是处于升序段,那后半段一定有peak,反之亦然。 class solution: #A: integer array #return: index of any peak positions def findPeak(self,A): if not A: return -1 start = 0 end = le...
1publicclassSolution {2publicintfindPeakElement(int[] nums) {3//Binary Search to find peak4intstart = 0, end = nums.length - 1, mid = 0, prev = 0, next = 0;5while(start + 1 <end) {6mid = (end - start) / 2 +start;7prev = mid - 1;8next = mid + 1;9if(nums[mid]...
首先分析若使用传统的二分搜索,若A[mid] > A[mid - 1] && A[mid] < A[mid + 1],则找到一个peak为A[mid];若A[mid - 1] > A[mid],则A[mid]左侧必定存在一个peak,可用反证法证明:若左侧不存在peak,则A[mid]左侧元素必满足A[0] > A[1] > … > A[mid -1] > A[mid],与已知A[0]...
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。
162 Find Peak Element 思路: 暴力法: 每個值也search Binary Search: 當nums[mid] > nums[mid +1] , 左邊 nums[0] 至nums[mid] 必然存在peak element 別人代碼:1 Binary Search: class Solution { public int findPeakElement(int[] nums) {
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! 写出了 O(N) 的解法。 My code: public class Solution { public int findPeakElement(int[] nums) { if (nums == null || nums.length == 0) return...
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[ele_mid])returnpeakElementSearch(arr1,ele_low,(ele_mid-1),n);elsereturnpeakElementSearch(arr1,(ele_mid+1),ele_high,n);}// Function to find a peak element in an arrayintPeakFinding(intarr1[],intn){returnpeakElementSearch(arr1,0,n-1,n);}intmain(){intarr1[]={5,12,13,20,...
则media为peak element的情况就被忽略了 另外,头尾添加负无穷构造新数组时要考虑溢出。 另另外:num.push_back(-2147483647-1);//不能直接push -2147483648 ,若这样,自动识别-2147483648为无符号的,不能将-号应用于无符号类型 #pragma once#include<iostream>#include<vector>using namespace std;//取大值int max...