题目:
Suppose a sorted array is rotated at some pivot unknown to you beforehand.
(i.e., 0 1 2 4 5 6 7might become4 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.
解答代码:
int binarySearch(int A[],int start,int end,int key) { int mid; while (start <= end) { mid = (start + end) / 2; if (A[mid] == key) { return mid; } if (A[start] <= A[mid]) { if (A[start] <= key && key < A[mid]) end = mid; else start = mid + 1; } else { if (A[mid] < key && key <= A[end]) start = mid + 1; else end = mid; } } return -1; }
解答要点:找到mid后分为两半处理,其中必然有一半是有序的,判断key是否在这有序数组中,如果在则继续在这数组中查找,否则在另一数组中查找