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leetcode - Search in Rotated Sorted Array
Suppose a sorted array is rotated at some pivot unknown to you beforehand.
(i.e., 0 1 2 4 5 6 7
might become 4 5 6 7 0 1 2
).
You are given a target value to search. If found in the array return its index, otherwise return -1.
You may assume no duplicate exists in the array.
//利用二分,但是,这里要特殊处理,将数组分成两部分二分。时间复杂度O(logn) class Solution { public: int search(int A[], int n, int target) { return dfs(A,0,n-1,target); } private: int dfs(int A[], int bgn, int end, int target) { if(bgn <= end) { int mid = bgn + (end - bgn) / 2; if(A[mid] == target) { return mid; } if(A[bgn] <= A[mid]) { if(A[bgn] <= target && target < A[mid]) { dfs(A,bgn,mid-1,target); } else { dfs(A,mid+1,end,target); } } else { if(A[mid] < target && target <= A[end]) { dfs(A,mid+1,end,target); } else { dfs(A,bgn,mid-1,target); } } } else { return -1; } } };
leetcode - Search in Rotated Sorted Array
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