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

Posted by 111qqz on Thursday, April 13, 2017

TOC

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;
    }

};

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