LeetCode 34. Search for a Range

题目：https://leetcode.com/problems/search-for-a-range/
思路：看LogN级别时间复杂度的要求就应该可以想到要用二分查找类似的做，关键问题是有重复元素，就是说如果找到了a[i]==target，还得顾及一下他的左右邻居的值是不是也一样的，如果是的话，那说明还得继续找
代码：

class Solution {
public:
    int searchMin(vector<int>& nums, int target) {
        int l = 0;
        int r = nums.size() - 1;
        int m;
        while (l <= r) {
            m = (l + r)/2;
            if(nums[m] == target){
                if(m == 0)
                    return m;
                else if(nums[m-1] < target) //targst cannot be in the left part
                    return m;
                else {
                    r = m - 1;
                }
            } else if(nums[m] < target) {
                l = m + 1;
            } else {
                r = m - 1;
            }
        }
        return -1;
    }
    int searchMax(vector<int>& nums, int target) {
        int l=0;
        int r = nums.size() - 1;
        int m;
        while(l <= r) {
            m = (l + r)/2;
            if(nums[m] == target){
                if (m == r)
                    return m;
                else if(nums[m+1] > target)
                    return m;
                else
                    l = m + 1;
            } else if (nums[m] < target) {
                l = m + 1;
            } else {
                r = m - 1;
            }
        }
        return -1;
    }
    vector<int> searchRange(vector<int>& nums, int target) {
        vector<int> result;
        result.push_back(searchMin(nums, target));
        result.push_back(searchMax(nums, target));
        return result;
    }
};