Bulls and Cows (E)

題目

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.

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.

Please note that both secret number and friend's guess may contain duplicate digits.

Example 1:

Input: secret = "1807", guess = "7810"

Output: "1A3B"

Explanation: 1 bull and 3 cows. The bull is 8, the cows are 0, 1 and 7.

Example 2:

Input: secret = "1123", guess = "0111"

Output: "1A1B"

Explanation: The 1st 1 in friend's guess is a bull, the 2nd or 3rd 1 is a cow.

Note: You may assume that the secret number and your friend's guess only contain digits, and their lengths are always equal.

題意

猜數字游戲，回傳"aAbB"形式的反饋資訊，

思路

兩次回圈：第一次記錄secret中所有數字的個數，第二次找位置正確的bulls同時，找guess數字在secret中出現的個數total，total - bulls就是cows，

一次回圈：如果對應字符相等，則bulls++；如果不相等，若secret對應字符出現次數為負數，說明該字符之前在guess中出現過，則cows++，若guess對應字符出現次數為正數，說明該字符之前在secret中出現過，則cows++，

代碼實作

Java

兩次回圈

class Solution {
    public String getHint(String secret, String guess) {
        int[] nums = new int[10];
        for (char c : secret.toCharArray()) {
            nums[c - '0']++;
        }

        int A = 0, total = 0;
        for (int i = 0; i < secret.length(); i++) {
            if (secret.charAt(i) == guess.charAt(i)) {
                A++;
            }
            if (nums[guess.charAt(i) - '0'] != 0) {
                nums[guess.charAt(i) - '0']--;
                total++;
            }
        }

        return A + "A" + (total - A) + "B";
    }
}

一次回圈

class Solution {
    public String getHint(String secret, String guess) {
        int nums[] = new int[10];
        int A = 0, B = 0;

        for (int i = 0; i < secret.length(); i++) {
            char s = secret.charAt(i);
            char g = guess.charAt(i);

            if (s == g) {
                A++;
            } else {
                if (nums[s - '0'] < 0) B++;
                if (nums[g - '0'] > 0) B++;
                nums[s - '0']++;
                nums[g - '0']--;
            }
        }

        return A + "A" + B + "B";
    }
}

JavaScript

兩次回圈

/**
 * @param {string} secret
 * @param {string} guess
 * @return {string}
 */
var getHint = function (secret, guess) {
  let A = 0
  let hash = new Array(10).fill(0)
  for (let i = 0; i < secret.length; i++) {
    hash[parseInt(secret[i])]++
    if (secret[i] === guess[i]) A++
  }
  let total = 0
  for (let c of guess) {
    let index = parseInt(c)
    if (hash[index] > 0) {
      hash[index]--
      total++
    }
  }
  return `${A}A${total - A}B`
}

一次回圈

/**
 * @param {string} secret
 * @param {string} guess
 * @return {string}
 */
var getHint = function (secret, guess) {
  let A = 0
  let B = 0
  let hash = new Array(10).fill(0)
  for (let i = 0; i < secret.length; i++) {
    if (secret[i] === guess[i]) {
      A++
    } else {
      if (hash[secret[i]]++ < 0) B++
      if (hash[guess[i]]-- > 0) B++
    }
  }
  return `${A}A${B}B`
}

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