Search in Rotated Sorted Array II
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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：一个已经排序好的数组。被按某个位置旋转了一次，给定一个值target，在该旋转后的数组里查找该值。（数组中的元素可能反复）
思路：二分查找
难点在于确定往数组的哪一半段继续二分查找
设起点、中间点、终点分别为 start、middle、end (採用前闭后开的区间表示方法
假设target = A[middle] return middle
假设A[middle] > A[start]，则[start,middle)单调递增
1.假设target < A[middle] && target >= A[start]，则 end = middle
2.start = middle + 1, otherwise
假设A[middle] < A[start]，则[middle,end)单调递增
1.假设target > A[middle] && target <= A[end - 1]，则 start = middle + 1
2.end = middle, otherwise
假设A[middle] == A[start]。那A[start]也不会是target，能够通过start++; 去掉A[start]
复杂度：时间O(n)，空间O(1)
int search(int A[], int n, int target){
int start = 0, end = n, middle ;
while(start < end){
middle = (start + end) / 2;
if(A[middle] == target) return middle;
if(A[middle] > A[start]){
if(target >= A[start] && target < A[middle]){
end = middle;
}else{
start = middle + 1;
}
}else if(A[middle] < A[start]){
if(target > A[middle] && target <= A[end - 1]){
start = middle + 1;
}else{
end = middle;
}
}
else{
++start;
}
}
return -1;
}