【LeetCode】34. Search for a Range

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].

题意：找出已排序数组中指定元素的起止位置。

思路：因为要求时间复杂度为O(logn)，所以用二分法

　　1.利用二分法确定指定元素的起始位置，同时确定该数组中是否存在该元素

　　2.利用二分法确定元素的终止位置

class Solution {
public:
    vector<int> searchRange(vector<int>& nums, int target) {
    
    int left=0,right=nums.size()-1;
    int mid;
    vector<int> res(2,-1);
    if(!nums.size()) return res;
    while(left<right){
        mid=(left+right)/2;
        if(nums[mid]<target) left=mid+1;
        else right=mid;
    }
    if(nums[left]!=target) return res;
    else res[0]=left;
    right = nums.size()-1;
    while(left<right){
        mid=(left+right)/2+1;                   //注意！！
        if(nums[mid]>target) right=mid-1;
        else left=mid;
    }
    res[1]=right;
    return res;
    }
};