[LeetCode] Find Peak Element, leetcodepeak

[LeetCode] Find Peak Element, leetcodepeak

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 in 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 shocould return the index number 2.

Note:
Your solution shocould be in logarithmic complexity.

This question requires that the data of an adjacent element is not equal, find the local maximum value, and return the corresponding subscript. Note that the time complexity of O (log N) is met.

If you do not consider the time complexity constraints, this question is quite simple. Starting from the first element, if it is greater than the subsequent element, num [0] is one of the local maximum values, and returns the subscript 0; otherwise, it will continue to traverse until an element is greater than its adjacent element.

However, considering the time complexity, we naturally think of binary search. If the intermediate element is greater than the adjacent element, the subscript of the intermediate element is returned. Otherwise, the side greater than the intermediate element must have a local largest element.

The continuation of the binary search, but I am very wordy, And I will post a more concise method later.

Int findPeakElement (const vector <int> & num) {int len = num. size (); if (! Len) exit (1); if (len = 1) return 0; if (len = 2) {if (num [0]> num [1]) return 0; else return 1;} int left = 0; int right = len-1; int mid = (left + right)/2; while (num [mid] <max (num [mid-1], num [mid + 1]) {if (num [mid] <num [mid-1]) {right = mid;} else {left = mid;} mid = (left + right)/2; if (mid = 0 | mid = len-1) return mid;} // brief int FindPeakElement (int num [], int len) {int l = 0; int r = len-1; int mid = (l + r)/2; while (l <r) {if (num [mid] <num [mid + 1]) {l = mid + 1 ;} else r = mid; mid = (l + r)/2;} return l ;}