1. 題目

給定一個(gè)非空字符串，其中包含字母順序打亂的英文單詞表示的數(shù)字0-9。按升序輸出原始的數(shù)字。

注意:
輸入只包含小寫英文字母。
輸入保證合法并可以轉(zhuǎn)換為原始的數(shù)字，這意味著像 “abc” 或 “zerone” 的輸入是不允許的。
輸入字符串的長(zhǎng)度小于 50,000。

示例 1: 輸入: "owoztneoer" 輸出: "012" (zeroonetwo)示例 2: 輸入: "fviefuro" 輸出: "45" (fourfive)

來(lái)源：力扣（LeetCode）
鏈接：https://leetcode-cn.com/problems/reconstruct-original-digits-from-english
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2. 解題

• 先找出只唯一擁有某字符的單詞
• Zero，tWo，foUr，siX，eiGht
• 然后上面找完了，再找剩下的唯一擁有字符的
• One，Three，Five，Seven
• 最后留下 nine

class Solution { public:string originalDigits(string s) {int count[26] = {0};for(auto& ch : s)count[ch-'a']++;int num[10] = {0};num[0] = count['z'-'a'];count['z'-'a'] -= num[0];count['e'-'a'] -= num[0];count['r'-'a'] -= num[0];count['o'-'a'] -= num[0];num[2] = count['w'-'a'];count['t'-'a'] -= num[2];count['w'-'a'] -= num[2];count['o'-'a'] -= num[2];num[4] = count['u'-'a'];count['f'-'a'] -= num[4];count['o'-'a'] -= num[4];count['u'-'a'] -= num[4];count['r'-'a'] -= num[4];num[6] = count['x'-'a'];count['s'-'a'] -= num[6];count['i'-'a'] -= num[6];count['x'-'a'] -= num[6];num[8] = count['g'-'a'];count['e'-'a'] -= num[8];count['i'-'a'] -= num[8];count['g'-'a'] -= num[8];count['h'-'a'] -= num[8];count['t'-'a'] -= num[8];num[1] = count['o'-'a'];count['o'-'a'] -= num[1];count['n'-'a'] -= num[1];count['e'-'a'] -= num[1];num[3] = count['t'-'a'];count['t'-'a'] -= num[3];count['h'-'a'] -= num[3];count['r'-'a'] -= num[3];count['e'-'a'] -= 2*num[3];num[5] = count['f'-'a'];count['f'-'a'] -= num[5];count['i'-'a'] -= num[5];count['v'-'a'] -= num[5];count['e'-'a'] -= num[5];num[7] = count['s'-'a'];count['s'-'a'] -= num[7];count['e'-'a'] -= 2*num[7];count['v'-'a'] -= num[7];count['n'-'a'] -= num[7];num[9] = count['i'-'a'];string ans;for(int i = 0; i < 10; ++i){while(num[i]--)ans += to_string(i);}return ans;} };

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