[LeetCode] #34 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.

本题利用二分算法的思想，判断target是在前半段还是后半段，再分开查找。时间：7ms。代码如下：

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