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class Solution {
    public String largestWordCount(String[] messages, String[] senders) {
        Map<String, Integer> senderWordCount = new HashMap<>();
        String maxSender = "zzzzzzzzzzz";
        int maxCount = 0;
        for (int i = 0; i < messages.length; i++) {
            senderWordCount.put(senders[i], senderWordCount.getOrDefault(senders[i], 0) + getWordCount(messages[i]));
            if (senderWordCount.get(senders[i]) > maxCount
                    || ((senderWordCount.get(senders[i]) == maxCount) && senders[i].compareTo(maxSender) > 0)) {
                maxCount = senderWordCount.get(senders[i]);
                maxSender = senders[i];
            }
        }
        return maxSender;
    }

    private int getWordCount(String message) {
        return message.split(" ").length;
    }
}

用一个变量存maxSender是谁, 一个变量存maxCount是多少即可.

时间复杂度: O(n)
空间复杂度: O(n) 因为用了HashMap.