LeetCode OJ--Search for a Range

http://oj.leetcode.com/problems/search-for-a-range/

要求复杂度为O(lgn)，用二分查找的思想。

#include <iostream>
#include <vector>
using namespace std;

class Solution {
public:
    void fun(int* A,int start,int end,int target,vector<int> &ans)
    {
        if(start>end || ( start == end && A[start]!= target ))
        {
            ans.push_back(-1);
            ans.push_back(-1);
            return;
        }
        if(start == end && A[start] == target)//只有一个元素了
        {
            ans.push_back(start);
            return;
        }
        if(A[start] == target && target ==A[end])
        {
            ans.push_back(start);
            ans.push_back(end);
            return;
        }
        int middle = (start + end)/2;
        if(A[middle]<target)
            fun(A,middle+1,end,target,ans);
        else if(A[middle]>target)
            fun(A,start,middle-1,target,ans);
        else
        {
            ans.push_back(middle);
            return;
        }
        return;
    }
    vector<int> searchRange(int A[], int n, int target) {
        vector<int> ans;
        if(n == 0)
            return ans;

        fun(A,0,n-1,target,ans);

        int small,large;
        if(ans[0] == -1) //肯定没找到
        {
            ans.clear();
            ans.push_back(-1);
            ans.push_back(-1);
            return ans;
        }
        if(ans.size()==1) //ans中只有一个元素
            small = large = ans[0];

        if(ans.size() == 2)
        {
            small = ans[0];
            large = ans[1];
        }
        while(small>=0 && A[small] == target) //往左边扩展
            small--;
        while(large<=n-1 && A[large] == target) //往右边扩展
            large++;
        if(small == -1 || A[small]!=target) //考虑上面while循环的退出条件
            small++;
        if(large== n || A[large]!=target)
            large--;
        ans.resize(2); //可能ans的size是1
        ans[0] = small; //将扩展后的再赋值回来
        ans[1] = large;
        return ans;
    }
};

int main()
{
    Solution myS;
    int A[] = {0,0,2,3,4,4,4,5};
    vector<int> ans = myS.searchRange(A,8,0);
    return 0;
}