leetcode34. Search for a Range

问题描述：

Given an array of integers nums 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].

Example 1:

Input: nums = [5,7,7,8,8,10], target = 8
Output: [3,4]

Example 2:

Input: nums = [5,7,7,8,8,10], target = 6
Output: [-1,-1]

 

问题思路：

这道题要求我们找到target在数组中的下标范围，若不存在则返回[-1, -1], 给出的数组为递增数组

考察我们如何判定该坐标为范围开始坐标以及范围结束坐标

注意边界：当target出现在最后一个位置的时候

 

代码：

class Solution {
public:
    vector<int> searchRange(vector<int>& nums, int target) {
        vector<int> ret(2,-1);
        for(int i = 0; i < nums.size(); i++){
            if(nums[i] == target){
                if(ret[0] == -1){
                    ret[0] = i;
                }
                continue;
            }
            if(nums[i] != target && ret[0] != -1 && ret[1] == -1)
                ret[1] = i-1;
        }
        if(ret[0] != -1 && ret[1] == -1){
            ret[1] = nums.size() - 1;
        }
        return ret;
    }
};