leetCode题解之寻找一个数在有序数组中的范围Search for a Range

1、问题描述

Given an array of integers sorted in ascending order, find the starting and ending position of a given target value.

Your algorithm's runtime complexity must be in the order of O(log n).

If the target is not found in the array, return [-1, -1].

For example,
Given [5, 7, 7, 8, 8, 10] and target value 8,
return [3, 4].

给定一个按照上升排序的数组和一个数，找出这个数在该数组中的开始位置和结束位置。如果该数不存在数组中。返回[-1,-1].

2、问题分析

题目要求在log(N)的时间内。选择二分查找，现在数组中找出该数的一个位置，然后向左右两边扫描，找出左边的位置和右边的位置。

3、代码

vector<int> searchRange(vector<int>& nums, int target) {
        // 数组是有序的 ，先用二分查找找出一个目标数。然后向两边寻找
        
        vector<int> ans;
        
        if(nums.size() == 0)
        {
            ans.push_back(-1);
            ans.push_back(-1);
            return ans;
        }
        
        int left = 0,right = nums.size()-1;
        int index = -1;
        while(left <= right)
        {
            int mid = left + (right - left)/2;
            if(nums[mid] == target)
            {
                index = mid;
                break;
            }
                
            else if( nums[mid] > target )
                right = mid -1;
            else
                left = mid + 1;
        }
        
        // target doesn't exist in array
        if( index == -1)
        {
            ans.push_back(-1);
            ans.push_back(-1);
            return ans;
        }
         
        // target exist in array
        
        int indexL  = index;
        int indexR = index;
        while( nums[indexL] == target  && indexL >= 0 ) indexL--;
        while( nums[indexR] == target  && indexR < nums.size() ) indexR++;
        
        int n = nums.size();
        ans.push_back(indexL+1);
        ans.push_back(indexR-1);
        
        return ans;