Search in Rotated Sorted Array I && II

对翻过一次的排序数组二分查找，要利用好已排序这个条件

class Solution {public:    int search(int A[], int n, int target) {        int left = 0, right = n-1;        while(left <= right){             int mid = (left+right)/2;             if(A[mid] == target)               return mid;             if(A[left] <= A[mid]){              if(A[left] <= target && target < A[mid])                right = mid-1;              else                 left = mid+1;             }             else {              if(A[mid] < target && target <= A[right])                left = mid+1;              else                 right = mid-1;             }        }        return -1;    }};

II中允许重复数字，碰到相等的向前移。

class Solution {public:    bool search(int A[], int n, int target) {        int left = 0, right = n-1;        while(left <= right){             int mid = (left+right)/2;             if(A[mid] == target)               return true;             if(A[left] < A[mid]){              if(A[left] <= target && target < A[mid])                right = mid-1;              else                 left = mid+1;             }             else if(A[left] > A[mid]){              if(A[mid] < target && target <= A[right])                left = mid+1;              else                 right = mid-1;             }             else left++;        }        return false;    }};