subject

Ideas

1. The topic seems quite simple It is demand. O(log n) Complexity The first reaction is dichotomy
2. The answer is exquisite , To learn

Code

        // Dichotomy seems to be able to find   Have a try 
        // Set boundaries 
        int start = 0;
        int end= nums.length - 1;
        while (start<end){
    
            //middle Doing so can avoid and start or end equal 
            int middle = start + (end - start) / 2;
            if (nums[middle] > nums[middle + 1]) {
     // The peak must be bigger than its next 
                end = middle;
            } else {
     // Add 1 It must be possible to narrow the scope 
                start = middle + 1;
            }
        }
        return start;