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
AC代码
class Solution {
public:
int search(vector& nums, int target) {
if (nums.empty()) return -1;
int ans = -1, left = 0, right = nums.size() - 1, mid;
while (left <= right) {
mid = (left + right) / 2;
if (nums[mid] == target)
return mid;
else if (nums[mid] >= nums[left]) { // mid左边有序
if (target >= nums[left] && target < nums[mid])
right = mid - 1;
else if (target >= nums[left] && target > nums[mid])
left = mid + 1;
else //target不在左半边
left = mid + 1;
}
else if (nums[mid] <= nums[right]) { // mid右边有序
if (target >= nums[mid] && target <= nums[right])
left = mid + 1;
else //要么target不在右半边，要么在右半边同时target AC代码（稍稍精简）
class Solution {
public:
int search(vector& nums, int target) {
if (nums.empty()) return -1;
int ans = -1, left = 0, right = nums.size() - 1, mid;
while (left <= right) {
mid = (left + right) / 2;
if (nums[mid] == target) return mid;
else if (nums[mid] >= nums[left]) { // mid左边有序
if (target >= nums[left] && target < nums[mid]) right = mid - 1;
else left = mid + 1;
}
else { // mid右边有序
if (target >= nums[mid] && target <= nums[right]) left = mid + 1;
else right = mid - 1;
}
}
return ans;
}
};
总结
1、两分钟O(n)，两小时O(logn)，边界问题折磨的要死，后来发现不需要单独考虑1个元素与2个元素的情况
2、二分法过于不熟悉，边界值全靠试错，需多加练习