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php-abstract-class-and-interface

发表于 2016-11-10 更新于 2020-01-12 分类于 php 阅读次数： Disqus：

PHP抽象类和接口

抽象类与接口
抽象类内可以包含非抽象函数，即可实现函数
抽象类内必须包含至少一个抽象方法，抽象类和接口均不能实例化
抽象类可以设置访问级别，接口默认都是public
类可以实现多个接口但不能继承多个抽象类
类必须实现抽象类和接口里的抽象方法，不一定要实现抽象类的非抽象方法
接口内不能定义变量，但是可以定义常量

示例代码

<?php
interface int1{
    const INTER1 = 111;
    function inter1();
}
interface int2{
    const INTER1 = 222;
    function inter2();
}
abstract class abst1{
    public function abstr1(){
        echo 1111;
    }
    abstract function abstra1(){
        echo 'ahahahha';
    }
}
abstract class abst2{
    public function abstr2(){
        echo 1111;
    }
    abstract function abstra2();
}
class normal1 extends abst1{
    protected function abstr2(){
        echo 222;
    }
}

result

1
2
3

PHP Fatal error:  Abstract function abst1::abstra1() cannot contain body in new.php on line 17

Fatal error: Abstract function abst1::abstra1() cannot contain body in php on line 17

summary-ranges-228

发表于 2016-10-12 更新于 2020-01-12 分类于 leetcode 阅读次数： Disqus：

problem

Given a sorted integer array without duplicates, return the summary of its ranges.

For example, given [0,1,2,4,5,7], return ["0->2","4->5","7"].

题解

每一个区间的起点nums[i]加上j是否等于nums[i+j]
参考

Code

class Solution {
public:
    vector<string> summaryRanges(vector<int>& nums) {
        int i = 0, j = 1, n;
        vector<string> res;
        n = nums.size();
        while(i < n){
            j = 1;
            while(j < n && nums[i+j] - nums[i] == j) j++;
            res.push_back(j <= 1 ? to_string(nums[i]) : to_string(nums[i]) + "->" + to_string(nums[i + j - 1]));
            i += j;
        }
        return res;
    }
};

minimum-size-subarray-sum-209

发表于 2016-10-11 更新于 2020-01-12 分类于 leetcode 阅读次数： Disqus：

problem

Given an array of n positive integers and a positive integer s, find the minimal length of a subarray of which the sum ≥ s. If there isn’t one, return 0 instead.

For example, given the array [2,3,1,2,4,3] and s = 7,
the subarray [4,3] has the minimal length under the problem constraint.

题解

参考，滑动窗口，跟之前Data Structure课上的online算法有点像，链接

Code

class Solution {
public:
    int minSubArrayLen(int s, vector<int>& nums) {
        int len = nums.size();
        if(len == 0) return 0;
        int minlen = INT_MAX;
        int sum = 0;
        
        int left = 0;
        int right = -1;
        while(right < len)
        {
            while(sum < s && right < len)
                sum += nums[++right];
            if(sum >= s)
            {
                minlen = minlen < right - left + 1 ? minlen : right - left + 1;
                sum -= nums[left++];
            }
        }
        return minlen > len ? 0 : minlen;
    }
};

binary-watch

发表于 2016-09-29 更新于 2020-01-12 分类于 leetcode 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:

1 2	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;
    }
};

34_Search_for_a_Range

发表于 2016-08-14 更新于 2020-01-12 分类于 leetcode 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;
    }
};