[LeetCode] Search in Rotated Sorted Array II [36]
称号
Follow up for "Search in Rotated Sorted Array":
What if duplicates are allowed?
Would this affect the run-time complexity? How and why?
Write a function to determine if a given target is in the array.
原题链接(点我)解题思路
这题和Search in Rotated Sorted Array问题类似。只是这个题同意元素反复。假设有元素反复,此时採取最保守的策略。一次缩小一个范围。
代码实现
class Solution { public: bool search(int A[], int n, int target) { if(A==NULL || n<=0) return false; int begin = 0, end = n-1, mid = 0; while(begin<=end){ mid = (begin+end)/2; if(A[mid] == target) return true; if(A[mid] < A[end]){ //后半段有序 if(A[end]>=target && A[mid]<target) begin = mid+1; else end = mid - 1; }else if(A[mid] > A[end]){ //前半段有序 if(A[begin]<=target && A[mid] > target) end = mid-1; else begin = mid+1; }else // 最保守策略,缩小一个范围 --end; } return false; } };
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