Leetcode 33. Search in Rotated Sorted Array

题目链接

问题描述

Suppose an array sorted in ascending order 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.

Your algorithm's runtime complexity must be in the order of O(log n).

Example 1:

Input: nums = [4,5,6,7,0,1,2], target = 0
Output: 4

Example 2:

Input: nums = [4,5,6,7,0,1,2], target = 3
Output: -1

解题思路

针对循环递增数组，以数组中间元素将整个数组分成两部分：当中间元素大于首元素时，前半部分为严格递增数组，后半部分为循环递增数组；当中间元素小于末尾元素时，前半部分为循环递增数组，后半部分为严格递增数组。

代码如下：

class Solution {
public:
	int search(vector<int>& nums, int target) {
		if (nums.size() == 0) return -1;
		int start = 0, end = nums.size() - 1;
		while (start <= end) {
			int mid = (end - start) / 2 + start;
			if (target == nums[mid]) return mid;
			if (nums[mid] < target) {
				if (nums[end] > nums[mid]) {
					if (nums[end] == target) return end;
					if (nums[end] > target) {
						start = mid + 1;
					}
					else {
						end = mid - 1;
					}
				}
				else {
					start = mid + 1;
				}
			}
			else {
				if (nums[end] > nums[mid]) {
					end = mid - 1;
				}
				else {
					if (nums[start] == target) return start;
					if (nums[start] > target) {
						start = mid + 1;
					}
					else {
						end = mid - 1;
					}
				}
			}
		}
		return -1;
	}
};