Search in Rotated Sorted Array

Suppose a sorted array 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.

Analyse: Binary search. 

Runtime: 4ms. 

class Solution {
public:
    int findMin(vector<int>& nums) {
        if(nums.empty()) return INT_MAX;
        
        int lowIndex = 0, highIndex = nums.size() - 1;
        while(lowIndex < highIndex) {
            int midIndex = lowIndex + (highIndex - lowIndex) / 2;
            
            if(nums[midIndex] > nums[highIndex])
                lowIndex = midIndex + 1;
            else
                highIndex = midIndex;
        }
        return nums[lowIndex];
    }
};

分析：用二分查找法。mid = (low + high) / 2。考虑如下两种情形：

1. 如果nums[low] <= nums[mid]表示这一段区间是递增的，倘若target的值在两者之间，那么确定上界为mid - 1，否则下界为mid + 1；

2. 如果nums[low] > nums[mid]表示mid在最大值之后（举例： [4， 5， 6，0， 1， 2， 3]。数字0在6之后）如果target在nums[mid]和nums[high]之间，则可确定下界为mid + 1;否则上界为mid - 1。

运行时间7ms

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