162. Find Peak Element

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

Given an input array nums, where nums[i] ≠ nums[i+1], find a peak element and return its index.

The array may contain multiple peaks, in that case return the index to any one of the peaks is fine.

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

Example 1:

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

Example 2:

Input: nums = [1,2,1,3,5,6,4]
Output: 1 or 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.

Note:

Your solution should be in logarithmic complexity.

普通方法：

class Solution {
    public int findPeakElement(int[] nums) {
        if (nums.length == 1) return 0;
        int[] newNums = new int[nums.length + 2];
        System.arraycopy(nums, 0, newNums, 1, nums.length);
        newNums[0] = Integer.MIN_VALUE;
        newNums[newNums.length - 1] = Integer.MIN_VALUE;
        for (int i = 1; i < newNums.length - 1; ++i) {
            if (newNums[i] > newNums[i - 1] && newNums[i] > newNums[i + 1]) return i - 1;
        }
        return -1;
    }
}

文艺方法：

Recursive Binary Search https://leetcode.com/problems/find-peak-element/solution/

public class Solution {
    public int findPeakElement(int[] nums) {
        return search(nums, 0, nums.length - 1);
    }
    public int search(int[] nums, int l, int r) {
        if (l == r)
            return l;
        int mid = (l + r) / 2;
        if (nums[mid] > nums[mid + 1])
            return search(nums, l, mid);
        return search(nums, mid + 1, r);
    }
}

二分查找：

public class Solution {
    public int findPeakElement(int[] nums) {
        int left = 0, right = nums.length - 1;
        while (left < right) {
            int mid = left + (right - left) / 2;
            if (nums[mid] < nums[mid + 1]) left = mid + 1;
            else right = mid;
        }
        return right;
    }
}

 

二逼方法：

class Solution {
    public int findPeakElement(int[] nums) {
         for (int i = 0; i < nums.length - 1; i++) {
            if (nums[i] > nums[i + 1])
                return i;
        }
        return nums.length - 1;
    }
}

为什么可以这样呢？

因为只要找到局部peak就满足题意了。

 

class Solution {
    public int findPeakElement(int[] nums) {
        if(nums.length == 1) return 0;
        int len = nums.length;
        int left = 0;
        int right = len - 1;
        int mid = 0;
        if(nums[len-1] > nums[len-2]) return len-1;
        if(nums[0] > nums[1]) return 0;
        while(left <= right){
            mid = left + (right - left)/2;
            //System.out.println(mid);
            if(nums[mid] < nums[mid+1]){
                left = mid + 1;
            }else{
                right = mid - 1;
            }
        }
        return left;

    }
}

二分查找，有个奇怪的地方，peak既然是大于邻居的，corner case就可以先判断两边。。

不行的话再二分查找，mid ＜ mid + 1，说明答案可能在右边，left = mid + 1. 最后返回left