文章目录
codewars-js练习
2021/1/24
github 地址
【1】<8kyu>【Sum The Strings】
Create a function that takes 2 positive integers in form of a string as an input, and outputs the sum (also as a string):
example：
sumStr("4", "5") // should output "9"
sumStr("34", "5") // should output "39"
solution：
<script type="text/javascript">
function sumStr(a,b) {
var sum = +a + +b;
console.log(sum);
return sum.toString();
}
// 验证
console.log(sumStr("4", "5") ); // '9'
</script>
【2】<6kyu>【Write Number in Expanded Form】
You will be given a number and you will need to return it as a string in Expanded Form.
example：
expandedForm(12); // Should return '10 + 2'
expandedForm(42); // Should return '40 + 2'
expandedForm(70304); // Should return '70000 + 300 + 4'
思路：
- 先将num切割放入到数组中；
- 遍历数组并进行运算，然后将非0的数字放入到新的数组中；
- 最后将新数组中的数据通过+进行连接
solution：
<script type="text/javascript">
function expandedForm(num) {
// 将num切割放入到数组中
var arr = num.toString().split('');
// console.log(arr);
var length = arr.length;
var newArr = [];
for(var i=0;i<length;i++){
var str = arr[i] * Math.pow(10,length-i-1);
// 将非0的数字放入到新的数组中
if(str !=0){
newArr.push(str);
}
}
// 将新数组中的数据通过+连接
// console.log(newArr.join(' + '));
return newArr.join(' + ');
}
// 验证
console.log(expandedForm(12)); // Should return '10 + 2'
console.log(expandedForm(42)); // Should return '40 + 2'
console.log(expandedForm(70304)); // Should return '70000 + 300+4'
</script>
【3】<7kyu>【String ends with?】
Complete the solution so that it returns true if the first argument(string) passed in ends with the 2nd argument (also a string).
example：
solution('abc', 'bc') // returns true
solution('abc', 'd') // returns false
solution：
<script type="text/javascript">
function solution(str, ending){
// 用endsWith判断字符串是否以谁结束
var temp = str.endsWith(ending);
return temp;
}
// 验证
console.log(solution('abc', 'bc')); // true
console.log(solution('abc', 'd')); // false
console.log(solution('abcde', 'abc'))//false
</script>
知识点
- startsWith(anotherString)
- 判断是否以为anotherString开头
- endsWith(anotherString)
- 判断是否以anotherString为结尾
"abcd".startsWith("ab"); // true
"abcd".startsWith("bc"); // false
"abcd".endsWith("cd"); // true
"abcd".endsWith("e"); // false
"a".startsWith("a"); // true
"a".endsWith("a"); // true
【4】<6kyu>【Delete occurrences of an element if it occurs more than n times】
Given a list lst and a number N, create a new list that contains each number of lst at most N times without reordering. For example if N = 2, and the input is [1,2,3,1,2,1,2,3], you take [1,2,3,1,2], drop the next [1,2] since this would lead to 1 and 2 being in the result 3 times, and then take 3, which leads to [1,2,3,1,2,3].
example：
deleteNth ([1,1,1,1],2) // return [1,1]
deleteNth ([20,37,20,21],1) // return [20,37,21]
solution：
<script type="text/javascript">
function deleteNth(arr,n){
// 判断数组是否为空
if(!arr){
return null;}
// 判断当n<1时，则说明1次都不能出现，所以要返回[]
if(n < 1){
return [];}
var result = [];
var itemCounts = {
};
// 遍历数组 判断每个元素出现的次数
for(var i=0;i<arr.length;i++){
var item = arr[i];
var count = itemCounts[item] ||0;
// 如果出现的次数<n，就将该元素放入新的数组
if(count < n){
result.push(item);
itemCounts[item] = count+1;
}
}
// console.log(result);
return result;
}
// 验证
console.log(deleteNth ([1,1,1,1],2)); // [1,1]
console.log(deleteNth ([20,37,20,21],1)); //[20,37,21]
</script>
【5】<6kyu>【Vasya - Clerk】
The new “Avengers” movie has just been released! There are a lot of people at the cinema box office standing in a huge line. Each of them has a single 100, 50 or 25 dollar bill. An “Avengers” ticket costs 25 dollars.
Vasya is currently working as a clerk. He wants to sell a ticket to every single person in this line.
Can Vasya sell a ticket to every person and give change if he initially has no money and sells the tickets strictly in the order people queue?
Return YES, if Vasya can sell a ticket to every person and give change with the bills he has at hand at that moment. Otherwise return NO.
example：
tickets([25, 25, 50]) // => YES
tickets([25, 100]) // => NO. Vasya will not have enough money to give change to 100 dollars
tickets([25, 25, 50, 50, 100]) // => NO. Vasya will not have the right bills to give 75 dollars of change (you can't make two bills of 25 from one of 50)
思路
- 如果是25，就收下，所以n25++;；
- 如果是50，也收下，但要找25，所以n50+; n25–;；
- 如果是100，收下，先找25，再判断n50是否大于0，如果还有50则优先找50，没有就找2张25，所以n25–; n50>0?n50–:n25-=2;；
solution：
<script type="text/javascript">
function tickets(peopleInLine){
// console.log(peopleInLine);
var len = peopleInLine.length;
var [n25,n50,n100] = [0,0,0];
for(let v of peopleInLine){
if(v == 25) n25++;
if(v == 50) {
n50++; n25--;}
if(v == 100){
n25--; n50>0?n50--:n25-=2;}
if(n25<0 || n50<0) return 'NO';
// console.log(n25,n50,n100);
}
return 'YES';
}
// 验证
console.log(tickets([25, 25, 50, 50])); // YES
console.log(tickets([25, 100])); //NO
console.log(tickets([25, 25, 50, 50, 100]));//NO
</script>
以上为自己思路供大家参考，可能有更优的思路。