LeetCode_ Binary search_ Medium_ 162. looking for peaks

1. subject

The peak element refers to the element whose value is strictly greater than the left and right adjacent values .

Give you an array of integers nums, Find the peak element and return its index . The array may contain multiple peaks , under these circumstances , Return to any peak position .

You can assume nums[-1] = nums[n] = -∞ .

You must achieve a time complexity of O(log n) Algorithm to solve this problem .

Example 1：

Input ：nums = [1,2,3,1]
Output ：2
explain ：3 Is the peak element , Your function should return its index 2.

Example 2：
Input ：nums = [1,2,1,3,5,6,4]
Output ：1 or 5
explain ： Your function can return the index 1, Its peak element is 2; Or return index 5, Its peak element is 6.

Tips ：
1 <= nums.length <= 1000
-2³¹ <= nums[i] <= 2³¹ - 1
For all that works i There are nums[i] != nums[i + 1]

source ： Power button （LeetCode）
link ：https://leetcode.cn/problems/find-peak-element

2. Ideas

（1） Find the maximum （ But it does not meet the time complexity requirements ）
Because the title guarantees for all valid i There are nums[i] != nums[i + 1], So array nums The elements on both sides of the maximum must be strictly smaller than the maximum itself . therefore , The position of the maximum value is a feasible peak position . So just do the array nums Do a walk , Find the subscript corresponding to the maximum value . However, the time complexity of this method is O(n), Does not meet the requirements of the topic .

（2） Iterative climbing
Train of thought reference Official solution to this problem .

（3） Ideas 2 Binary search optimization based on
Train of thought reference Official solution to this problem .

3. Code implementation （Java）

// Ideas 1———— Find the maximum （ But it does not meet the time complexity requirements ）
class Solution {
    
    public int findPeakElement(int[] nums) {
    
        int index = 0;
        for (int i = 0; i < nums.length; i++) {
    
            if (nums[i] > nums[index]) {
    
                index = i;
            }
        }
        return index;
    }
}

// Ideas 2———— Iterative climbing 
class Solution {
    
    public int findPeakElement(int[] nums) {
    
        int length = nums.length;
        //Math.random(): Randomly select greater than or equal to  0.0  And less than  1.0  Pseudorandom of  double  value 
        int index = (int) (Math.random() * length);
        while (!(compare(nums, index - 1, index) < 0 && compare(nums, index, index + 1) > 0)) {
    
            if (compare(nums, index, index + 1) < 0) {
    
                index += 1;
            } else {
    
                index -= 1;
            }
        }
        return index;
    }
    
    /*  Auxiliary function , Input subscript  i, Returns a tuple  (0/1, nums[i])  Easy to handle  nums[-1]  as well as  nums[n]  The situation of the border  */
    public int[] get(int[] nums, int index) {
    
        if (index == -1 || index == nums.length) {
    
            // Boundary situation 
            return new int[]{
    0, 0};
        } else {
    
            // Normal condition 
            return new int[]{
    1, nums[index]};
        }
    }
    
    public int compare(int[] nums, int index1, int index2) {
    
        int[] num1 = get(nums, index1);
        int[] num2 = get(nums, index2);
        if (num1[0] != num2[0]) {
    
            return num1[0] > num2[0] ? 1 : -1;
        }
        if (num1[1] == num2[1]) {
    
            return 0;
        }
        return num1[1] > num2[1] ? 1 : -1;
    }
}

// Ideas 3———— Ideas 2 Binary search optimization based on 
class Solution {
    
    public int findPeakElement(int[] nums) {
    
        int length = nums.length;
        int left = 0, right = length - 1, res = -1;
        while (left <= right) {
    
            int mid = (left + right) / 2;
            if (compare(nums, mid - 1, mid) < 0 && compare(nums, mid, mid + 1) > 0) {
    
                res = mid;
                break;
            }
            if (compare(nums, mid, mid + 1) < 0) {
    
                left = mid + 1;
            } else {
    
                right = mid - 1;
            }
        }
        return res;
    }

    public int[] get(int[] nums, int idx) {
    
        if (idx == -1 || idx == nums.length) {
    
            return new int[]{
    0, 0};
        }
        return new int[]{
    1, nums[idx]};
    }

    public int compare(int[] nums, int idx1, int idx2) {
    
        int[] num1 = get(nums, idx1);
        int[] num2 = get(nums, idx2);
        if (num1[0] != num2[0]) {
    
            return num1[0] > num2[0] ? 1 : -1;
        }
        if (num1[1] == num2[1]) {
    
            return 0;
        }
        return num1[1] > num2[1] ? 1 : -1;
    }
}