本文介绍了找出数组元素出现的次数的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！
问题描述
我是 JavaScript 新手，我已经学习和练习了大约 3 个月，希望我能在这个主题上得到一些帮助.我正在制作一个扑克游戏，我想要做的是确定我是否有一对、两对、三对、四对或满屋.
I am new to JavaScript, I have been learning and practicing for about 3 months and hope I can get some help on this topic. I'm making a poker game and what I'm trying to do is determine whether i have a pair, two pairs, three of a kind, four of a kind or a full house.
比如在[1, 2, 3, 4, 4, 4, 3]中，1出现一次，4出现三次，以此类推.
For instance, in [1, 2, 3, 4, 4, 4, 3], 1 appears one time, 4 appears three times, and so on.
我怎么可能让我的电脑告诉我一个数组元素出现了多少次?
How could I possibly ask my computer to tell me how many times an array element appears?
解决了，这是最终产品.
Solved, here's the final product.
<script type="text/javascript">
var deck = [];
var cards = [];
var convertedcards = [];
var kinds = [];
var phase = 1;
var displaycard = [];
var options = 0;
var endgame = false;
// Fill Deck //
for(i = 0; i < 52; i++){
deck[deck.length] = i;
}
// Distribute Cards //
for(i = 0; i < 7; i++){
cards[cards.length] = Number(Math.floor(Math.random() * 52));
if(deck.indexOf(cards[cards.length - 1]) === -1){
cards.splice(cards.length - 1, cards.length);
i = i - 1;
}else{
deck[cards[cards.length - 1]] = "|";
}
}
// Convert Cards //
for(i = 0; i < 7; i++){
convertedcards[i] = (cards[i] % 13) + 1;
}
// Cards Kind //
for(i = 0; i < 7; i++){
if(cards[i] < 13){
kinds[kinds.length] = "H";
}else if(cards[i] < 27 && cards[i] > 12){
kinds[kinds.length] = "C";
}else if(cards[i] < 40 && cards[i] > 26){
kinds[kinds.length] = "D";
}else{
kinds[kinds.length] = "S";
}
}
// Card Display //
for(i = 0; i < 7; i++){
displaycard[i] = convertedcards[i] + kinds[i];
}
// Hand Strenght //
var handstrenght = function(){
var usedcards = [];
var count = 0;
var pairs = [];
for(i = 0, a = 1; i < 7; a++){
if(convertedcards[i] === convertedcards[a] && a < 7 && usedcards[i] != "|"){
pairs[pairs.length] = convertedcards[i];
usedcards[a] = "|";
}else if(a > 6){
i = i + 1;
a = i;
}
}
// Flush >.< //
var flush = false;
for(i = 0, a = 1; i < 7; i++, a++){
if(kinds[i] === kinds[a] && kinds[i] != undefined){
count++;
if(a >= 6 && count >= 5){
flush = true;
count = 0;
}else if(a >= 6 && count < 5){
count = 0;
}
}
}
// Straight >.< //
var straight = false;
convertedcards = convertedcards.sort(function(a,b){return a-b});
if(convertedcards[2] > 10 && convertedcards[3] > 10 && convertedcards[4] > 10){
convertedcards[0] = 14;
convertedcards = convertedcards.sort(function(a,b){return a-b});
}
alert(convertedcards);
if(convertedcards[0] + 1 === convertedcards[1] && convertedcards[1] + 1 === convertedcards[2] && convertedcards[2] + 1 === convertedcards[3] && convertedcards[3] + 1 === convertedcards[4]){
straight = true;
}else if(convertedcards[1] + 1 === convertedcards[2] && convertedcards[2] + 1 === convertedcards[3] && convertedcards[3] + 1 === convertedcards[4] && convertedcards[4] + 1 === convertedcards[5]){
straight = true;
}else if(convertedcards[2] + 1 === convertedcards[3] && convertedcards[3] + 1 === convertedcards[4] && convertedcards[4] + 1 === convertedcards[5] && convertedcards[5] + 1 === convertedcards[6]){
straight = true;
}
// Royal Flush, Straight Flush, Flush, Straight >.< //
var royalflush = false;
if(straight === true && flush === true && convertedcards[6] === 14){
royalflush = true;
alert("You have a Royal Flush");
}
else if(straight === true && flush === true && royalflush === false){
alert("You have a straight flush");
}else if(straight === true && flush === false){
alert("You have a straight");
}else if(straight === false && flush === true){
alert("You have a flush");
}
// Full House >.< //
if(pairs[0] === pairs[1] && pairs[1] != pairs[2] && pairs.length >= 3){
fullhouse = true;
alert("You have a fullhouse");
}else if(pairs[0] != pairs[1] && pairs[1] === pairs[2] && pairs.length >= 3){
fullhouse = true;
alert("You have a fullhouse");
}else if(pairs[0] != pairs[1] && pairs[1] != pairs[2] && pairs[2] === pairs[3] && pairs.length >= 3){
fullhouse = true;
alert("You have a fullhouse");
}
// Four of a kind >.< //
else if(pairs[0] === pairs[1] && pairs[1] === pairs[2] && pairs.length > 0){
alert("You have four of a kind");
}
// Three of a kind >.< //
else if(pairs[0] === pairs[1] && flush === false && straight === false && pairs.length === 2){
alert("You have three of a kind");
}
// Double Pair >.< //
else if(pairs[0] != pairs[1] && flush === false && straight === false && pairs.length > 1){
alert("You have a double pair");
}
// Pair >.< //
else if(pairs.length === 1 && flush === false && straight === false && pairs.length === 1 ){
alert("You have a pair");
}
alert(pairs);
};
while(endgame === false){
if(phase === 1){
options = Number(prompt("Your hand: " + displaycard[0] + " " + displaycard[1] + "\n\n" + "1. Check" + "\n" + "2. Fold"));
}else if(phase === 2){
options = Number(prompt("Your hand: " + displaycard[0] + " " + displaycard[1] + "\n\n" + displaycard[2] + " " + displaycard[3] + " " + displaycard[4] + "\n\n" + "1. Check" + "\n" + "2. Fold"));
}else if(phase === 3){
options = Number(prompt("Your hand: " + displaycard[0] + " " + displaycard[1] + "\n\n" + displaycard[2] + " " + displaycard[3] + " " + displaycard[4] + " " + displaycard[5] + "\n\n" + "1. Check" + "\n" + "2. Fold"));
}else if(phase === 4){
options = Number(prompt("Your hand: " + displaycard[0] + " " + displaycard[1] + "\n\n" + displaycard[2] + " " + displaycard[3] + " " + displaycard[4] + " " + displaycard[5] + " " + displaycard[6] + "\n\n" + "1. Check" + "\n" + "2. Fold"));
}
switch(options){
case 1:
if(phase === 5){
handstrenght();
endgame = true;
}else{
phase++;
}
break;
case 2:
endgame = true;
break;
default:
endgame = true;
break;
}
}
</script>
推荐答案
- 为总数保留一个变量
- 循环遍历数组并检查当前值是否与您要查找的值相同，如果是，则将总计数加一
- 循环后，总计数包含您要查找的数字在数组中的次数
显示您的代码，我们可以帮助您找出问题所在
Show your code and we can help you figure out where it went wrong
这是一个简单的实现(因为您没有不起作用的代码)
Here's a simple implementation (since you don't have the code that didn't work)
var list = [2, 1, 4, 2, 1, 1, 4, 5];
function countInArray(array, what) {
var count = 0;
for (var i = 0; i < array.length; i++) {
if (array[i] === what) {
count++;
}
}
return count;
}
countInArray(list, 2); // returns 2
countInArray(list, 1); // returns 3
countInArray 也可以实现为
countInArray could also have been implemented as
function countInArray(array, what) {
return array.filter(item => item == what).length;
}
更优雅，但可能没有那么高，因为它必须创建一个新数组.
More elegant, but maybe not as performant since it has to create a new array.
这篇关于找出数组元素出现的次数的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！