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社区首页 >专栏 >Leetcode 题目解析之 Bulls and Cows

Leetcode 题目解析之 Bulls and Cows

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ruochen
发布2022-01-10 20:14:43
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发布2022-01-10 20:14:43
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文章被收录于专栏:若尘的技术专栏

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.

代码语言:txt
复制
    public String getHint(String secret, String guess) {
        if (secret == null || guess == null
                || secret.length() != guess.length()) {
            return "";
        }
        char[] s = secret.toCharArray();
        char[] g = guess.toCharArray();
        int A = 0, B = 0, n = s.length;
        // 统计bulls
        for (int i = 0; i < n; i++) {
            if (s[i] == g[i]) {
                s[i] = '#';
                g[i] = '#';
                A++;
            }
        }
        // 统计cows
        Map<Character, Integer> map = new HashMap<Character, Integer>();
        for (int i = 0; i < n; i++) {
            if (s[i] != '#') {
                if (!map.containsKey(s[i])) {
                    map.put(s[i], 1);
                } else {
                    int times = map.get(s[i]);
                    times++;
                    map.put(s[i], times);
                }
            }
        }
        for (int i = 0; i < n; i++) {
            if (g[i] != '#') {
                if (map.containsKey(g[i]) && map.get(g[i]) != 0) {
                    int times = map.get(g[i]);
                    times--;
                    map.put(g[i], times);
                    B++;
                }
            }
        }
        return A + "A" + B + "B";
    }

原创声明:本文系作者授权腾讯云开发者社区发表,未经许可,不得转载。

如有侵权,请联系 cloudcommunity@tencent.com 删除。

原创声明:本文系作者授权腾讯云开发者社区发表,未经许可,不得转载。

如有侵权,请联系 cloudcommunity@tencent.com 删除。

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