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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.
题意：一个已经排序过得有重复元素的数组翻转后查找target。思路：每次判断A[left],A[mid],A[right]的关系，然后找到确定是递增的部分接着二分，也有特殊情况就是left++,right--
class Solution {
public:
bool search(int A[], int n, int target) {
int left = 0, right = n-1;
while (left <= right) {
int mid = left + (right - left) / 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[mid] < A[right]) {
if (A[mid] < target && target <= A[right])
left = mid + 1;
else right = mid - 1;
} else if (A[left] == A[mid]) left += 1;
else if (A[right] == A[mid]) right -= 1;
}
return false;
}
};
LeetCode Search in Rotated Sorted Array II
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原文地址：http://blog.csdn.net/u011345136/article/details/44975453
