1.Linux创建用户并给每一个用户创建的默认目录是/bash
Linux强制结束进程的kill命令的参数是 kill -9 进程
2.canvas标签有width
和height
两个属性,style设置样式
- eg:
<canvas id="canvas" width="500px" height="500px" style="border:1px solid blue"></canvas>
3.H5中用来获取当前位置的经纬度API是:navigator.geolocation.getCurrentPosition
https://blog.youkuaiyun.com/Jioho_chen/article/details/83592630?ops_request_misc=%257B%2522request%255Fid%2522%253A%2522161639599516780357293600%2522%252C%2522scm%2522%253A%252220140713.130102334…%2522%257D&request_id=161639599516780357293600&biz_id=0&utm_medium=distribute.pc_search_result.none-task-blog-2allsobaiduend~default-1-83592630.pc_search_result_no_baidu_js&utm_term=h5%E8%8E%B7%E5%8F%96%E5%BD%93%E5%89%8D%E4%BD%8D%E7%BD%AE%E7%BB%8F%E7%BA%AC%E5%BA%A6
4.MySQL中锁按粒度分类
- 表级锁
- 表共享读锁
- 表独占写锁
- 行级锁
- 共享锁
- 排它锁
- 页级锁
5.线程与进程
6.资源间有序访问的实现是同步
7.求字符串子串是回文字符串的个数
- 暴力解决法
var countSubstrings = function(s) {
var len = s.length;
var left, right,maxArr = [],maxLength = 0;
for(var i = 0; i < len; i++) {
for(var j = i+1; j < len; j++) {
left = i;
right = j;
var flag = false;
maxLength = 0;
while(left <= right) {
flag = false;
var m = left == right?1:2;
if(s[left] == s[right]) {
maxLength += m;
flag = true;
left++;
right--;
}else {
break;
}
}
if(flag == true) {
maxArr.push(maxLength);
}
}
}
return len + maxArr.length;//最长回文子串则是maxArr中最大值,需要先push(1),可能是一个字符是最长回文子串
};
- 动态规划法
作者:xiao_ben_zhu
链接:https://leetcode-cn.com/problems/palindromic-substrings/solution/shou-hua-tu-jie-dong-tai-gui-hua-si-lu-by-hyj8/
来源:力扣(LeetCode)
著作权归作者所有。商业转载请联系作者获得授权,非商业转载请注明出处。
const countSubstrings = (s) => {
const len = s.length;
let count = 0;
const dp = new Array(len);
for (let j = 0; j < len; j++) {
for (let i = 0; i <= j; i++) {
if (s[i] == s[j] && (j - i <= 1 || dp[i + 1])) {
dp[i] = true;
count++;
} else {
dp[i] = false;
}
}
}
return count;
};