Leetcode-33:Search in Rotated Sorted Array

2018-01-29

问题描述：

Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.

>
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.

问题示例

1	(i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2).

我的答案

int search(vector<int>& nums, int target) {
    int start = 0;
    int end = nums.size() - 1;
    int mid;
    while (start<=end) {
        mid = (start + end)/2;
        //处理边界
        if(end - start < 2){
            for (int i = start; i<=end; i++) {
                if(nums[i] == target){
                    return i;
                }
            }
            return  -1;
        }
        if(nums[mid]>nums[start]){
            if (target == nums[mid]) {
                return mid;
            }else if(target < nums[mid]&&target>=nums[start]){
                end = mid - 1;
            }else{
                start = mid + 1;
            }
        }else{
            if (target == nums[mid]) {
                return mid;
            }else if(target > nums[mid]&&target<=nums[end]){
                start = mid + 1;
            }else{
                end = mid - 1;
            }
        }
    }
    return -1;
}

其他高效答案

int search(vector<int>& nums, int target) {
        int start = 0, end = nums.size() - 1;
        if(nums.size() == 0) return -1;
        while(start <= end){
            int mid = start + (end - start) / 2;
            if(nums[mid] == target) return mid;
            if(nums[mid] > nums[end]){ // there exists rotation; the middle element is in the left part of the array
                if((target >= nums[start]) && (target <= nums[mid])){
                    end = mid - 1;
                }
                else start = mid + 1;
            }
            else{//there exists rotation; the middle element is in the right part of the array
                if((target >= nums[mid]) && (target <= nums[end])){
                    start = mid + 1;
                }
                else end = mid - 1;
            }
        }
        return -1;
        
        }