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What hurts more, the pain of hard work or the pain of regret?

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发表于 分类于 阅读次数： Disqus：

problem

A binary watch has 4 LEDs on the top which represent the hours (0-11), and the 6 LEDs on the bottom represent the minutes (0-59).

Each LED represents a zero or one, with the least significant bit on the right.

For example, the above binary watch reads “3:25”.

Given a non-negative integer n which represents the number of LEDs that are currently on, return all possible times the watch could represent.

Example:

Input: n = 1
Return: ["1:00", "2:00", "4:00", "8:00", "0:01", "0:02", "0:04", "0:08", "0:16", "0:32"]

Note:

  • The order of output does not matter.
  • The hour must not contain a leading zero, for example “01:00” is not valid, it should be “1:00”.
  • The minute must be consist of two digits and may contain a leading zero, for example “10:2” is not valid, it should be “10:02”.

题解

又是参（chao）考（xi）别人的代码，嗯，就是这么不要脸，链接

Code

class Solution {
public:
    vector<string> readBinaryWatch(int num) {
        vector<string> res;
        for (int h = 0; h < 12; ++h) {
            for (int m = 0; m < 60; ++m) {
                if (bitset<10>((h << 6) + m).count() == num) {
                    res.push_back(to_string(h) + (m < 10 ? ":0" : ":") + to_string(m));
                }
            }
        }
        return res;
    }
};

发表于 分类于 阅读次数： Disqus：

question

34. Search for a Range

Original Page

Given a sorted array of integers, 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].

analysis

一开始就想到了二分查找，但是原来做二分查找的时候一般都是找到确定的那个数就完成了，
这里的情况比较特殊，需要找到整个区间，所以需要两遍查找，并且一个是找到小于target
的最大索引，一个是找到大于target的最大索引，代码参考leetcode discuss,这位仁
兄也做了详细的分析解释。

code

class Solution {
public:
    vector<int> searchRange(vector<int>& nums, int target) {
        vector<int> ret(2, -1);
        int i = 0, j = nums.size() - 1;
        int mid;
        while(i < j){
            mid = (i + j) / 2;
            if(nums[mid] < target) i = mid + 1;
            else j = mid;
        }
        if(nums[i] != target) return ret;
        else {
            ret[0] = i;
            if((i+1) < (nums.size() - 1) && nums[i+1] > target){
                ret[1] = i;
                return ret;
            }
        }   //一点小优化
        j = nums.size() - 1;
        while(i < j){
            mid = (i + j) / 2 + 1;
            if(nums[mid] > target) j = mid - 1;
            else i = mid;
        }
        ret[1] = j;
        return ret;
    }
};

发表于 分类于 阅读次数： Disqus：

玩一下swoole的websocket

WebSocket是HTML5开始提供的一种在单个TCP连接上进行全双工通讯的协议。WebSocket通信协议于2011年被IETF定为标准RFC 6455，WebSocketAPI被W3C定为标准。
,在web私信，im等应用较多。背景和优缺点可以参看wiki。

环境准备

因为swoole官方还不支持windows，所以需要装下linux,之前都是用ubuntu，
这次就试一下centos7，还是满好看的，虽然虚拟机会默认最小安装，需要在安装
时自己选择带gnome的，当然最小安装也是可以的，只是最后需要改下防火墙。
然后是装下PHP，Nginx什么的，我是用Oneinstack，可以按需安装
给做这个的大大点个赞。

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发表于 分类于 阅读次数： Disqus：

Invert a binary tree

     4
   /   \
  2     7
 / \   / \
1   3 6   9

to

     4
   /   \
  7     2
 / \   / \
9   6 3   1

Trivia:
This problem was inspired by this original tweet by Max Howell:

Google: 90% of our engineers use the software you wrote (Homebrew),
but you can’t invert a binary tree on a whiteboard so fuck off.

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* invertTree(TreeNode* root) {
        if(root == NULL) return root;
        TreeNode* temp;
        temp = invertTree(root->left);
        root->left = invertTree(root->right);
        root->right = temp;
        return root;
    }
};