299. Bulls and Cows

题目：

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.

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.

链接： http://leetcode.com/problems/bulls-and-cows/

题解：

公牛和奶牛游戏。使用HashMap存下来secret里的字符和count，然后同时遍历secret和guess就可以了。最后还要遍历一次map把多加的cow减掉。

Time Complexity - O(n)， Space Complexity - O(n)

public class Solution {

    public String getHint(String secret, String guess) {

        if(secret == null || guess == null || secret.length() != guess.length()) {

            return "0A0B";

        }

        int bulls = 0, cows = 0;

        Map<Character, Integer> map = new HashMap<>();

        for(int i = 0; i < secret.length(); i++) {

            char c = secret.charAt(i);

            if(!map.containsKey(c)) {

                map.put(c, 1);

            } else {

                map.put(c, map.get(c) + 1);

            }

        }

        for(int i = 0; i < secret.length(); i++) {

            char sChar = secret.charAt(i);

            char gChar = guess.charAt(i);

            if(sChar == gChar) {

                bulls++;

                map.put(gChar, map.get(gChar) - 1);

            } else if(map.containsKey(gChar)) {

                cows++;

                map.put(gChar, map.get(gChar) - 1);

            }

        }

        for(char c : map.keySet()) {

            if(map.get(c) < 0) {

                cows += map.get(c);

            }

        }

        return String.valueOf(bulls) + "A" + String.valueOf(cows) + "B";

    }

}

二刷:

主要参考了Discuss里面的解。

我们可以用一个数组来存bulls和cows。用s和g来表示数组的数字值
当 s = g时，我们找到了bull， bulls++
否则我们要看
1. nums[g] > 0的话，说明当前guess的这个数字曾经出现在secret中，这是一个cow，我们cow++
2. 我们也要看是否nums[s] < 0，这个表明当前ssecret的数字曾经出现在guess中，这也是一个cow，我们还是cow++
3. 我们用正数记录下nums[s]为bull的一个位置，nums[s]++，我们也用负数记录下guess中出现过的数字，nums[g]--，

Java:

Time Complexity - O(n)， Space Complexity - O(1)

public class Solution {

    public String getHint(String secret, String guess) {

        if(secret == null || guess == null || secret.length() != guess.length()) {

            return "0A0B";

        }

        int[] nums = new int[10];

        for (int i = 0; i < secret.length(); i++) {

            int s = secret.charAt(i) - '0';

            int g = guess.charAt(i) - '0';

            if (s == g) {

                bulls++;

            } else {

                if (nums[s] < 0) {  // bulls can be counted as cows

                    cows++;

                }

                if (nums[g] > 0) {  // found num but in diff position

                    cows++;

                }

                nums[s]++;

                nums[g]--;

            }

        }

        return bulls + "A" + cows + "B";

    }

}

三刷:

延续了二刷的解法。主要使用一个count[]数组保存之前出现过的secret digits和guess digits。当前sDigit == gDigit时，bulls增加。否则，当count[sDigit] < 0时，说明之前出现在guess里，当count[gDigit] > 0时，说明之前出现在secret里，这两种情况都要分别增加cows。之后再记录i这个位置的改动count[sDigit]++， count[gDigit]--。最后返回结果。

Java:

public class Solution {

    public String getHint(String secret, String guess) {

        if (secret == null || guess == null || secret.length() != guess.length()) {

            return "0A0B";

        }

        int bulls = 0;

        int cows = 0;

        int[] count = new int[10];

        for (int i = 0; i < secret.length(); i++) {

            int sDigit = secret.charAt(i) - '0';

            int gDigit = guess.charAt(i) - '0';

            if (sDigit == gDigit) {

                bulls++;

            } else {

                if (count[sDigit] < 0) {

                    cows++;

                }

                if (count[gDigit] > 0) {

                    cows++;

                }

            }

            count[sDigit]++;

            count[gDigit]--;

        }

        return bulls + "A" + cows + "B";

    }

}

Update:

public class Solution {

    public String getHint(String secret, String guess) {

        if (secret == null || guess == null || secret.length() != guess.length()) return "0A0B";

        int bullsCount = 0, cowsCount = 0;

        int len = secret.length();

        int[] count = new int[10];

        for (int i = 0; i < len; i++) {

            int sc = secret.charAt(i) - '0';

            int gc = guess.charAt(i) - '0';

            if (sc == gc) {

                bullsCount++;

            } else {

                if (count[gc] > 0) cowsCount++;

                if (count[sc] < 0) cowsCount++;

                count[sc]++;

                count[gc]--;

            }

        }

        return bullsCount + "A" + cowsCount + "B";

    }

}

Reference:

https://leetcode.com/discuss/67031/one-pass-java-solution

巴特西

299. Bulls and Cows

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