leetcode 34. Search for a Range (二分，找到一段值为tar的区间)

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].

思路：二分。。。

我好像根本不会二分啊？？？

二分查找

/* ***********************************************
Author :111qqz
Created Time :2017年04月13日 星期四 10时49分02秒
File Name :34.cpp
************************************************ */
class Solution {

public:

    int n;
    vector<int>res;
    int bin(int l,int r,vector<int>& nums,int tar) //
    {
	while (l<=r)
	{
	    int mid = (l+r)>>1;
	    if (nums[mid]>=tar) r = mid - 1;
	    else l = mid + 1;
	}
	return nums[r+1]==tar?r+1:-1;//找到第一个等于tar的下标
    }
    int bin2(int l,int r,vector<int>& nums,int tar)
    {
	while (l<=r)
	{
	    int mid = (l+r)>>1;
	    if (nums[mid]>tar) r = mid-1;
	    else l = mid + 1;
	}
	return nums[l-1]==tar?l-1:-1; //找到最后一个等于tar的下标
    }
    vector<int> searchRange(vector<int>& nums, int target) {
        n = nums.size();
	if (n==0) return vector<int>(2,-1);
	int L = bin(0,n-1,nums,target);
	int R = bin2(0,n-1,nums,target);
	if (L>=n) L = -1; //对于只有一个数的情况，需要维护一下。
	if (R>=n) R = -1;
	res.push_back(L);
	res.push_back(R);
	return res;
    }

};