问题描述：

34. Find First and Last Position of Element in Sorted Array

Medium

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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]

题目大意就是给定一个排序的向量和目标值，返回目标值的起始点和结束点坐标。

解题思路：

与上一题一样，还是二分查找的思路，lower_bound用来找左界，upper_bound用来找右界。找左界的时候，当中间值和目标值target相同的时候，继续在左半部分查找，最后返回左边界。

找右界的时候，当中间值和目标值target相同的时候，继续在右半部分查找，最后返回左边界。

源码

class Solution {
public:
    vector<int> searchRange(vector<int>& nums, int target) {
        int low = lower_bound(nums, target);
        int high = upper_bound(nums, target);
        if (low == high) 
            return {-1, -1};
        else
            return {low, high - 1};
    }
    int lower_bound(vector<int>& nums, int target) {
        const int N = nums.size();
        // [l, r)
        int l = 0, r = N;
        while (l < r) {
            int mid = l + (r - l) / 2;
            if (nums[mid] >= target) {
                r = mid;
            } else  {
                l = mid + 1;
            }
        }
        return l;
    }
    int upper_bound(vector<int>& nums, int target) {
        const int N = nums.size();
        // [l, r)
        int l = 0, r = N;
        while (l < r) {
            int mid = l + (r - l) / 2;
            if (nums[mid] <= target) {
                l = mid + 1;
            } else {
                r = mid;
            }
        }
        return l;
    }
};

还有另一种解法，就是调用c++的库函数：

class Solution {
public:
    vector<int> searchRange(vector<int>& nums, int target) {
        auto low = lower_bound(nums.begin(), nums.end(), target);
        auto high = upper_bound(nums.begin(), nums.end(), target);
        if (low == high) return {-1, -1};
        return {low - nums.begin(), high - nums.begin() - 1};
    }
};