旋转数组的二分查找

2018/10/02 Offer 共 1083 字,约 4 分钟

第一种是无重复的,也就是说左边和右边不会相等

33.Search in Rotated Sorted Array

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

第二种是有重复的,可能nums[l]==nums[r],这个要注意。

如81.Search in Rotated Sorted Array II

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

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