2017-08-23 19:18:17 +0000   |     algorithm leetcode string array   |   Viewed times   |    

题目

Given an arbitrary ransom note string and another string containing letters from all the magazines, write a function that will return true if the ransom note can be constructed from the magazines ; otherwise, it will return false.

Each letter in the magazine string can only be used once in your ransom note.

Note: You may assume that both strings contain only lowercase letters.

canConstruct("a", "b") -> false
canConstruct("aa", "ab") -> false
canConstruct("aa", "aab") -> true

基本思路

朴素的做法,把字符串当成一个字符列表来处理。复杂度 mransomNote的长度,nmagazine的长度。

高级一点,可以把字符串统计成字符的频率表。可以用一个Map,也可以用一个int[26]就够了。复杂度

解法一:用列表

代码

class Solution {
    public boolean canConstruct(String ransomNote, String magazine) {
        List<Character> mgz = new LinkedList<>();
        for (int i = 0; i < magazine.length(); i++) {
            mgz.add(magazine.charAt(i));
        }
        for (int i = 0; i < ransomNote.length(); i++) {
            if (!mgz.remove((Character)ransomNote.charAt(i))) { return false; }
        }
        return true;
    }
}

结果

ransom-note-1

解法二:统计字符频率

代码

class Solution {
    public boolean canConstruct(String ransomNote, String magazine) {
        int[] freq = countFreq(magazine);
        for (int i = 0; i < ransomNote.length(); i++) {
            int offset = ransomNote.charAt(i) - 'a';
            if (freq[offset] > 0) {
                --freq[offset];
            } else {
                return false;
            }
        }
        return true;
    }
    private int[] countFreq(String magazine) {
        int[] freq = new int[26];
        for (int i = 0; i < magazine.length(); i++) {
            ++freq[magazine.charAt(i)-'a'];
        }
        return freq;
    }
}

结果

ransom-note-2