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  • 423 Reconstruct Original Digits from English 从英文中重建数字

    给定一个非空字符串,其中包含字母顺序打乱的英文单词表示的数字0-9。按升序输出原始的数字。
    注意:
        输入只包含小写英文字母。
        输入保证合法并可以转换为原始的数字,这意味着像 "abc" 或 "zerone" 的输入是不允许的。
        输入字符串的长度小于 50,000。
    示例 1:
    输入: "owoztneoer"
    输出: "012" (zeroonetwo)
    示例 2:
    输入: "fviefuro"
    输出: "45" (fourfive)
    详见:https://leetcode.com/problems/reconstruct-original-digits-from-english/description/

    C++:

    方法一:

    class Solution {
    public:
        string originalDigits(string s) {
            string res = "";
            vector<int> counts(128, 0), nums(10, 0);
            for (char c : s)
            {
                ++counts[c];
            }
            nums[0] = counts['z'];
            nums[2] = counts['w'];
            nums[4] = counts['u'];
            nums[6] = counts['x'];
            nums[8] = counts['g'];
            nums[1] = counts['o'] - nums[0] - nums[2] - nums[4];
            nums[3] = counts['h'] - nums[8];
            nums[5] = counts['f'] - nums[4];
            nums[7] = counts['s'] - nums[6];
            nums[9] = counts['i'] - nums[6] - nums[8] - nums[5];
            for (int i = 0; i < nums.size(); ++i) 
            {
                for (int j = 0; j < nums[i]; ++j)
                {
                    res += (i + '0');
                }
            }
            return res;
        }
    };
    

     方法二:

    class Solution {
    public:
        string originalDigits(string s) {
            string res = "";
            vector<string> words{"zero", "two", "four", "six", "eight", "one", "three", "five", "seven", "nine"};
            vector<int> nums{0, 2, 4, 6, 8, 1, 3, 5, 7, 9}, counts(26, 0);
            vector<char> chars{'z', 'w', 'u', 'x', 'g', 'o', 'h', 'f', 's', 'i'};
            for (char c : s)
            {
                ++counts[c - 'a'];
            }
            for (int i = 0; i < 10; ++i) 
            {
                int cnt = counts[chars[i] - 'a'];
                for (int j = 0; j < words[i].size(); ++j) 
                {
                    counts[words[i][j] - 'a'] -= cnt;
                }
                while (cnt--)
                {
                    res += (nums[i] + '0');
                }
            }
            sort(res.begin(), res.end());
            return res;
        }
    };
    

     参考:https://www.cnblogs.com/grandyang/p/5996239.html

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  • 原文地址:https://www.cnblogs.com/xidian2014/p/8857212.html
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