https://leetcode.com/problems/search-in-rotated-sorted-array/
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
代码：
class Solution {
public:
int search(vector<int>& nums, int target) {
int n = nums.size();
map<int, int> mp;
mp.clear();
for(int i = 0; i < n; i ++)
mp[nums[i]] = i;
sort(nums.begin(), nums.end());
int l = 0, r = n - 1;
int mid;
while(l <= r) {
mid = (r - l) / 2 + l;
if(nums[mid] == target) return mp[nums[mid]];
else if(nums[mid] > target) r = mid - 1;
else l = mid + 1;
}
return -1;
}
};
惊险！0.93%
class Solution {
public:
int search(vector<int>& nums, int target) {
int left = 0, right = nums.size() - 1;
while (left <= right) {
int mid = left + (right - left) / 2;
if (nums[mid] == target) return mid;
else if (nums[mid] < nums[right]) {
if (nums[mid] < target && nums[right] >= target) left = mid + 1;
else right = mid - 1;
} else {
if (nums[left] <= target && nums[mid] > target) right = mid - 1;
else left = mid + 1;
}
}
return -1;
}
};
#Leetcode# 33. Search in Rotated Sorted Array
原文：https://www.cnblogs.com/zlrrrr/p/10000473.html