Bm19 looking for peak

Given a length of n Array of nums, Please find the peak and return its index . The array may contain multiple peaks , under these circumstances , Just return to any location .

    public int findPeakElement (int[] nums) {
        // write code here
        if (nums == null || nums.length == 0){
            return -1;
        }
        
        int i = 0;
        
        while (i <= nums.length - 1){
            boolean left = false;
            boolean right = false;
            if (i - 1 < 0 || nums[i] > nums[i-1]){
                left = true;
            }
            if (i + 1 > nums.length - 1 || nums[i] > nums[i+1]){
                right = true;
            }
            if (left && right){
                return i;
            }
            i++;
        }
        
        return -1;
    }

notes ： Direct judgment nums[i] > nums[i-1] && nums[i] > nums[i+1] There may be only one or two cases left out . 

Additional conditions ：

1. The peak element refers to the element whose value is strictly greater than the left and right adjacent values . Strictly greater than means that there can be no equal to

2. hypothesis  nums[-1] = nums[n] = −∞

3. For all that works i There are nums[i] != nums[i + 1]

4. You can use O(logN) Time complexity to achieve this problem ？

The principle of using dichotomy is ：

Use dichotomy , If there is a certain location nums[i] < nums[i+1] when , Then continue to look right , There must be an extreme value on the right .

（ Because the left side mid~mid+1 An ascending sequence has been completed , that , There is only one descending order on the right , The extreme value appears , If the right side is the green value in the figure below , that mid+1 It's extreme value , If it's red , Keep looking right , Until the last element at the latest , It must be the limit value , Why? ？ All the way to the right , Indicates that the left side is incremented , and nums[n] = −∞, So it must be satisfied ） The same is true for the left side .

 

    public int findPeakElement (int[] nums) {
        // write code here
        if (nums == null || nums.length == 0){
            return -1;
        }
        
        int low = 0;
        int high = nums.length - 1;
        while (low < high){
            int mid = low + (high - low) / 2;
            if (nums[mid] < nums[mid + 1]){
                low = mid + 1;
            } else {
                high = mid;
            }
        }
        
        return low;
    }