33. Search in Rotated Sorted Array

最新推荐文章于 2020-08-05 08:57:53 发布

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本文探讨了在已排序数组发生旋转后进行目标值搜索的问题，通过改进二分查找方法解决此挑战，强调了数组边界处理及判断条件的重要性。

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Search in Rotated Sorted Array My Submissions Question
Total Accepted: 92936 Total Submissions: 310089 Difficulty: Hard
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.

Subscribe to see which companies asked this question

还是改了挺久的，二分查找的变形，数组边界处理要注意，判断条件也要注意

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

                else{
                    first=mid+1;
                    mid=(last+first)/2;
                }
            }
            else{
                if(nums[last-1]>=target && nums[mid]<target){
                    //之前一直wa的原因，nums[last]用作判断，越界
                    first=mid+1;
                    mid=(last+first)/2;
                }
                else{
                    last=mid;
                    mid=(last+first)/2;
                }
            }

        }
        return -1;
    }
};