问题描述

A peak element is an element that is strictly greater than its neighbors.

Given an integer array nums, find a peak element, and return its index. If the array contains multiple peaks, return the index to any of the peaks.

You may imagine that nums[-1] = nums[n] = -∞.

You must write an algorithm that runs in O(log n) time.

测试样例

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Input: nums = [1,2,3,1]
Output: 2
Explanation: 3 is a peak element and your function should return the index number 2.

说明

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Input: nums = [1,2,1,3,5,6,4]
Output: 5
Explanation: Your function can return either index number 1 where the peak element is 2, or index number 5 where the peak element is 6.

解题

思路

注意条件中的 nums[i] != nums[i + 1]，即相邻元素不同

再考虑到 nums[-1] = nums[n] = -∞

则整个数组必有一个峰值，其左右元素均比它小，不管存在多少个波峰，我们寻找该峰值（实际上为最大值）即可

补充：

  1. 时间复杂度 O(logn)

代码

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class Solution {
    public int findPeakElement(int[] nums) {
        // 注意条件中的 nums[i] != nums[i + 1]，即相邻元素不同
        // 再考虑到 nums[-1] = nums[n] = -无穷
        // 则整个数组必有一个峰值，寻找该峰值即可
        int l = 0, r = nums.length - 1;
        
        while(l < r) {
            int mid = (l + r) / 2;
            
            int value_m_l = mid == 0 ? Integer.MIN_VALUE : nums[mid - 1];
            int value_m_r = mid == nums.length - 1 ? Integer.MAX_VALUE : nums[mid + 1];
            
            // peak value
            if(value_m_l < nums[mid] && nums[mid] > value_m_r) {
                return mid;
            }
            // 右侧值更大，区间缩小（包含最大值）
            else if(nums[mid] < value_m_r) {
                l = mid + 1;
            }
            // 左侧值更大，区间缩小（包含最大值）
            else {
                r = mid;
            }
        }
        
        return l;
    }
}