299. Bulls and Cows


You are playing the following Bulls and Cows game with your friend:
You write down a number and ask your friend to guess what the number is. Each time your friend makes a guess, you provide a hint that indicates how many digits in said guess match your secret number exactly in both digit and position (called "bulls") and how many digits match the secret number but locate in the wrong position (called "cows"). Your friend will use successive guesses and hints to eventually derive the secret number.

For example:

Secret number:  "1807"
Friend's guess: "7810"
Hint: 1 bull and 3 cows. (The bull is 8, the cows are 0, 1 and 7.) Write a function to return a hint according to the secret number and friend's guess, use A to indicate the bulls and B to indicate the cows. In the above example, your function should return "1A3B".

Please note that both secret number and friend's guess may contain duplicate digits, for example:

Secret number:  "1123"
Friend's guess: "0111"
In this case, the 1st 1 in friend's guess is a bull, the 2nd or 3rd 1 is a cow, and your function should return "1A1B". You may assume that the secret number and your friend's guess only contain digits, and their lengths are always equal.


這題目敘述有點長,我記得以前這是以前堂哥跟我玩的(猜數字遊戲)[http://codepen.io/skyyen999/full/VedwqQ/],上面那一大串英文的意思是,這邊有一串隱藏的號碼(secret number),然後你朋友會猜一串號碼, 如果號碼數字與位置都對了,給一個bull,數字對但位置不對,給一個cow。


Secret number:  "1807"
Friend's guess: "7810"

Secret number:  "1123"
Friend's guess: "0111"
第二個範例,第二個1得到一個bull,Friend's guess 第三個1得到一個cow(比對secret number的第一個1),因此得到1A1B


  1. 先判斷有幾個bull(位置數字都一樣)
  2. 出現bull將secret number跟friend's guess這個位子上的數字移除
  3. 剩下的字串再來判斷有幾個cow


 * @param {string} secret
 * @param {string} guess
 * @return {string}
var getHint = function(secret, guess) {
    var bull = 0;
    var cow  = 0;

    // 儲存secret[n],guess[n]數字不同的元素
    var skeep = [];
    var gkeep = [];

    //先判斷位置數字都一樣,剩下的用sK, gK來儲存
    for(var i in guess){
        // 位置數字都一樣,bull++
        if(secret[i]  == guess[i]){
        } else {

    // 因為bull已經處理過,這邊只要gkeep內的元素出現在skeep內,代表就是一個cow
    for(var j in gkeep){
        var findIndex = skeep.indexOf(gkeep[j]);
        if(findIndex != -1){
            skeep[findIndex] = null;

    return bull+ "A" + cow + "B"

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