81. 搜索旋转排序数组 II

10111 和 1110111101 这种。此种情况下 nums[start] == nums[mid]，分不清到底是前面有序还是后面有序，此时 start++ 即可。相当于去掉一个重复的干扰项。

package main
import "fmt"
func search(nums []int, target int) bool {
    left :=0
    right :=len(nums)-1
    for left<=right{
        mid :=left+(right-left)>>1
        if nums[mid]==target{
            return true
        }else if nums[mid]==nums[left]{
            left++
        } else if nums[left]<nums[mid]{
            if nums[left]<=target&&target<nums[mid]{
                right =mid-1
            }else {
                left = mid+1
            }
        }else {
            if nums[mid]<target&&target<=nums[right]{
                left = mid+1
            }else {
                right = mid-1
            }
        }
    }
    return false
}
func main() {
    fmt.Println(search([]int{2,5,6,0,0,1,2},0))
    fmt.Println(search([]int{2,5,6,0,0,1,2},3))
    fmt.Println(search([]int{3,1},1))
}