go语言题解LeetCode1160拼写单词示例详解

输入：words = ["cat","bt","hat","tree"], chars = "atach"
输出：6
解释： 
可以形成字符串 "cat" 和 "hat"，所以答案是 3 + 3 = 6。

输入：words = ["hello","world","leetcode"], chars = "welldonehoneyr"
输出：10
解释：
可以形成字符串 "hello" 和 "world"，所以答案是 5 + 5 = 10。

class Solution {
    public int countCharacters(String[] words, String chars) {
        int n=chars.length();
        int[] arr=new int[26];
        for(int i=0;i<n;i++){
            arr[chars.charAt(i)-'a']++;
        }
        int res=0;
        for(String word:words){
            int[] arr1=new int[26];
            for(int j=0;j<word.length();j++){
                arr1[word.charAt(j)-'a']++;
            }
            for(int i=0;i<26;i++){
                if(arr1[i]<=arr[i]) {
                    if(i==25) res=res+word.length();
                }else break;
            }
        }
        return res;
    }
}

class Solution {
    public int countCharacters(String[] words, String chars) {
        int[] c = new int[26];
        for(char cc : chars.toCharArray()) {
            c[(int)(cc - 'a')] += 1;
        }
        int res = 0;
        a: for(String word : words) {
            int[] w = new int[26];
            for(char ww : word.toCharArray()) {
                w[(int)(ww - 'a')] += 1;
            }
            for(int i=0; i<26; i++) {
                if(w[i] > c[i]) {
                    continue a;
                }
            }
            res += word.length();
        }
        return res;
    }
}

class Solution {
public:
    int countCharacters(vector<string>& words, string chars) {
        // 记录累计的长度之和
        int res = 0;
        // 构建chars的映射表
        int cnt[26];
        memset(cnt, 0, sizeof(int)*26);
        for (char c : chars)
        {
            ++cnt[c-'a'];
        }
​​​​​​​        // 遍历words去构建每个映射表，然后比较即可
        int curr[26];
        for (string word : words)
        {
            memset(curr, 0, sizeof(int)*26);
            for (char c : word)
            {
                ++curr[c-'a'];
            }
            bool IsOK= true;
            for (int i = 0; i < 26; ++i)
            {
                if (curr[i] > cnt[i])
                {
                    IsOK = false;
                    break;
                }
            }
            if (IsOK)
            {
                res += word.size();
            }
        }
        return res;
    }
};