33.搜索旋转排序数组

就只想到了O(n)的，
想到的二分，

1. target
2. target>=nums[left],在左边有序搜素
3. 自然是错误的，于是搁置不得了之。

看了题解恍然大悟。
以mid分成左右两个数组。
其中至少一个是有序的，且范围已知

• 如果target在此范围内，二分即可
• 不在此范围内，搜索另一个数组

class Solution {
    public int search(int[] nums, int target) {
        return searchHelp(nums,target,0,nums.length-1);
    }
    private int searchHelp(int[] nums,int target,int left,int right){
        if(left>right) return -1;
        int mid=left+((right-left)>>1);
        if(nums[mid]==target) return mid;
        if(mid-1>=left&&nums[left]<=nums[mid-1]){
            //有序在左边，且目标值在里面
            if(target>=nums[left]&&target<=nums[mid-1]){
                right=mid-1;
                while (left<=right){
                    mid=left+(right-left)/2;
                    if(nums[mid]==target) return mid;
                    if(nums[mid]>target) right=mid-1;
                    else left=mid+1;
                }
            }//不在里面，搜索右边无序的。
            return searchHelp(nums,target,mid+1,right);

        }else {
            //有序在右边，且目标在里面
            if(mid+1<nums.length&&target<=nums[right]&&target>=nums[mid+1]){
                left=mid+1;
                while (left<=right){
                    mid=left+(right-left)/2;
                    if(nums[mid]==target) return mid;
                    if(nums[mid]>target) right=mid-1;
                    else left=mid+1;
                }
            }
            return searchHelp(nums,target,left,mid-1);

        }
    }
}

还有一种更简洁的写法

class Solution {
    public int search(int[] nums, int target) {
        int left=0;
        int right=nums.length-1;
        while (left<=right){
            int mid=left+((right-left)>>1);
            if(nums[mid]==target) return mid;
            if(mid-1>=left&&nums[left]<=nums[mid-1]){
                if(target<=nums[mid-1]&&target>=nums[left]){
                    right=mid-1;
                }else  left=mid+1;
            }else{
                if(mid+1<nums.length&&target>=nums[mid+1]&&target<=nums[right]){
                    left=mid+1;
                } else right=mid-1;
            }
        }
        return -1;
    }
}

关于判断

 if(mid-1>=left&&nums[left]<=nums[mid-1]){
 nums[left]<=nums[mid-1] 取等号，是为了考虑只有一个数的时候