R8-二分查找篇

等等，不是？O(logn)？那岂不是一次遍历数组都超时了hh.

二分查找+分类 

class Solution {
    public int search(int[] nums, int target) {
        if (nums==null||nums.length==0){
            return -1;
        }
        int start=0;
        int end=nums.length-1;
        int mid;
        while (start<=end){
            mid=(start+end)/2;
            if (target==nums[mid]){
                return mid;
            }
            //前半段有序
            if (nums[start]<=nums[mid]){
                if (target>=nums[start]&&target<nums[mid]){
                    end=mid-1;
                }else{
                    start=mid+1;
                }
            }
            //后半段有序
            else{
                if (target<=nums[end]&&target>nums[mid]){
                    start=mid+1;
                }else{
                    end=mid-1;
                }
            }
        }
        return -1;
    }
}

ps:

1.

2.java不能连续的等号或者大于小于号，python可以