LeetCode 33：搜索旋转排序数组（Search in Rotated Sorted Array）解法汇总

更多LeetCode题解
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

Solution

看到复杂度$O(log(n))$，就想到了二分查找法，此题是二分查找的一个变体，多加一些判断条件就好了。

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