基于springboot+vue+Mysql的生鲜交易系统

文章目录

• 题目描述
• 思路
• 复杂度
• Code

题目描述

思路

复杂度

时间复杂度:

空间复杂度:

Code

class Solution {
    /**
     * Search for a rotationally sorted array(Binary search)
     *
     * @param nums   Given array
     * @param target Number to be found
     * @return int
     */
    public int search(int[] nums, int target) {
        if (nums == null || nums.length < 1) {
            return -1;
        }
        int left = 0;
        int right = nums.length - 1;
        while (left <= right) {
            int mid = left + (right - left) / 2;
            //Found
            if (nums[mid] == target) {
                return mid;
                //Right-side interval order
            } else if (nums[mid] <= nums[right]) {
                //On the right
                if (target >= nums[mid] && target <= nums[right]) {
                    left = mid + 1;
                } else {
                    right = mid - 1;
                }
            } else { //Left-side interval order
                //On the left
                if (target < nums[mid] && target >= nums[left]) {
                    right = mid - 1;
                } else {
                    left = mid + 1;
                }
            }
        }
        return -1;
    }
}