CRCK array 1.4

原创于 2016-03-27 00:55:16 发布 · 216 阅读

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本文介绍了两种判断两个字符串是否为字谜(anagram)的方法：一种使用排序算法，时间复杂度为nlogn，空间复杂度为O(1)；另一种使用计数法，时间复杂度和空间复杂度均为O(n)。

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Write a method to decide if two strings are anagrams or not.

There are two ways to solve this problem.

<pre name="code" class="cpp">// use the sort algorithm in C++.
// Time complexity (nLgn), space o(1);
bool isAnagrams(string a, string b) {
    // if the length of string a and b is not equal. There is no chance for them to be anagram.
    if(a.size() != b.size()) return false;
    sort(a.begin(), a.end()); sort(b.begin(), b.end());
    if(strcmp(a, b) == 0) return true;
    return false;
}
// another way is to use space O(n), time O(n)
bool isAnagrams(string a, string b) {
    // check the length first.
    if(a.size() != b.size()) return false;
    vector<int> counts(256, 0); // use an array to remember the number of occurrences. 
    for(int i = 0; i < a.size(); ++i) {
        counts[a[i]]++;
        counts[b[i]]--;
    }
    for(int i = 0; i < 256; ++i) {
        if(counts[i] < 0) return false; // any negative means the occurrences of a and b is different.
    }
    return true;
}