Find Peak Element

45% Accepted

There is an integer array which has the following features:

The numbers in adjacent positions are different.
A[0] < A[1] && A[A.length - 2] > A[A.length - 1].
We define a position P is a peek if:

A[P] > A[P-1] && A[P] > A[P+1]
Find a peak element in this array. Return the index of the peak.

Have you met this question in a real interview? Yes
Example
Given [1, 2, 1, 3, 4, 5, 7, 6]
Return index 1 (which is number 2) or 6 (which is number 7)

Note

The array may contains multiple peeks, find any of them.

Challenge

Time complexity O(logN)

Tags Expand

• Binary Search
• Array

思路

• 划分成小问题
• 二分搜索
• 注意边界

优化前

• 可能有边界问题
• start + 1 start - 1 end + 1 end - 1都有可能出现问题

class Solution {
    /**
     * @param A: An integers array.
     * @return: return any of peek positions.
     */
    public int findPeak(int[] A) {
        // write your code here
        if (A == null || A.length <= 2 ) {
            return -1;
        }

        int start = 0;
        int end = A.length -1;

        while (start + 1 < end) {
            int mid = start + (end - start) / 2;
            if( (A[mid] > A[mid-1]) && (A[mid] > A[mid+1]) ){
                return mid;
            }else if(A[mid] < A[mid+1]){
                start = mid;
            }else if(A[mid] > A[mid+1]){
                end = mid;
            }
        }

        if ((A[start] > A[start + 1]) && (A[start] > A[start - 1])) {
            return start;
        } else if ((A[end] > A[end + 1]) && (A[end] > A[end - 1])) {
            return end;
        } else {
            return -1;
        }
    }
}

优化后

• 没有了边界问题
• 因为第一个数和最后一个数肯定不是peak element，所以start = 1 end = A.length - 2
• 判断的时候因为是peak 一定是start 与 end中的最大值

class Solution {
    /**
     * @param A: An integers array.
     * @return: return any of peek positions.
     */
    public int findPeak(int[] A) {
        // write your code here
        int start = 1, end = A.length-2; // 1.答案在之间，2.不会出界
        while(start + 1 <  end) {
            int mid = (start + end) / 2;
            if(A[mid] < A[mid - 1]) {
                end = mid;
            } else if(A[mid] < A[mid + 1]) {
                start = mid;
            } else {
                end = mid;
            }
        }
        return A[start] > A[end] ? start : end
    }
}