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].

法一：

类似使用二分法，分别找出target的起始和终点

https://leetcode.com/problems/search-for-a-range/discuss/

class Solution {
public:
    vector<int> searchRange(vector<int>& nums, int target) {
        vector<int> ind(2, -1);
        if(nums.capacity() == 0) return ind;
        int l = 0, r = nums.capacity() - 1;
        while(r > l){    //寻找左边缘
            int m = (r + l)/2;
            if(nums[m] < target) l = m + 1;
            else r = m;
        }
        if(nums[l] != target) return ind;
        ind[0] = l;
        r = nums.capacity() - 1;
        while(r > l){     //寻找右边缘
            int m = (r + l)/2 + 1;    //加一使m更接近右边缘
            if(nums[m] > target) r = m - 1;
            else l = m;
        }
        ind[1] = r;
        return ind;
    }
};

法二：

使用函数库：

vector<int> searchRange(vector<int>& nums, int target) {
    auto bounds = equal_range(nums.begin(), nums.end(), target);
    if (bounds.first == bounds.second)
        return {-1, -1};
    return {bounds.first - nums.begin(), bounds.second - nums.begin() - 1};
}