[刷题防痴呆] 0423 - 从英文中重建数字 (Reconst

题目地址

https://leetcode.com/problems/reconstruct-original-digits-from-english/

题目描述

423. Reconstruct Original Digits from English

Given a string s containing an out-of-order English representation of digits 0-9, return the digits in ascending order.

 

Example 1:

Input: s = "owoztneoer"
Output: "012"
Example 2:

Input: s = "fviefuro"
Output: "45"

思路

• hashmap统计出每一个字符的count.
• 对应的数字实际上是脑筋急转弯.
• z对应0, w对应2, u对应4, x对应6, g对应8. 单次且仅在这几个数出现, 因此偶数全部可以确定.
• 之后推理出单数.
• 3对应h的数量减去8的数量. 5对应f的数量减去4的数量, 7对应s的数量减去6的数量. 9对应i的数量减去5, 6, 8的数量. 1对应n的数量减去7, 和9的两倍数量.

关键点

代码

• 语言支持：Java

class Solution {
    public String originalDigits(String s) {
        Map<Character, Integer> map = new HashMap<>();
        for (char c: s.toCharArray()) {
            map.put(c, map.getOrDefault(c, 0) + 1);
        }

        int[] res = new int[10];
        res[0] = map.getOrDefault('z', 0);
        res[2] = map.getOrDefault('w', 0);
        res[4] = map.getOrDefault('u', 0);
        res[6] = map.getOrDefault('x', 0);
        res[8] = map.getOrDefault('g', 0);
        res[3] = map.getOrDefault('h', 0) - res[8];
        res[5] = map.getOrDefault('f', 0) - res[4];
        res[7] = map.getOrDefault('s', 0) - res[6];
        res[9] = map.getOrDefault('i', 0) - res[5] - res[6] - res[8];
        res[1] = map.getOrDefault('n', 0) - res[7] - 2 * res[9];

        StringBuffer sb = new StringBuffer();
        for (int i = 0; i < 10; i++) {
            for (int j = 0; j < res[i]; j++) {
                sb.append(i);
            }
        }

        return sb.toString();
    }
}