题目描述

• 看了题目以后想到啥？
  • 字符数量统计
  • 银行家算法逐个拆解
  • 建立数字 - 字符串的全局映射
    

思路 && 代码

• 抄答案了，采取了评论区三叶dalao的写法（不得不说，人家的处理写法是真的优雅），学习学习。

class Solution {static String[] ss = new String[]{"zero", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine"};static int[] priority = new int[]{0, 8, 6, 3, 2, 7, 5, 9, 4, 1};public String originalDigits(String s) { // part 1： 字符数量统计int[] counts = new int[26];for(char c : s.toCharArray()) counts[c - 'a']++;// part 2：获取值（这里的嵌套循环处理很棒）StringBuilder sb = new StringBuilder();for(int i : priority) { // 按照"0, 8, 6, 3, 2, 7, 5, 9, 4, 1"的序列来int k = Integer.MAX_VALUE;for(char c : ss[i].toCharArray()) k = Math.min(k, counts[c - 'a']); // 获取当前最多的可行值for(char c : ss[i].toCharArray()) counts[c - 'a'] -= k; // 逐个减去可行值while(k-- > 0) sb.append(i); // 添加对应个数的 ans}// final: 排序、格式处理char[] cs = sb.toString().toCharArray();Arrays.sort(cs);return String.valueOf(cs);}
}

• 无注释版

class Solution {static String[] ss = new String[]{"zero", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine"};static int[] priority = new int[]{0, 8, 6, 3, 2, 7, 5, 9, 4, 1};public String originalDigits(String s) { int[] counts = new int[26];for(char c : s.toCharArray()) counts[c - 'a']++;StringBuilder sb = new StringBuilder();for(int i : priority) { int k = Integer.MAX_VALUE;for(char c : ss[i].toCharArray()) k = Math.min(k, counts[c - 'a']); for(char c : ss[i].toCharArray()) counts[c - 'a'] -= k; while(k-- > 0) sb.append(i); }char[] cs = sb.toString().toCharArray();Arrays.sort(cs);return String.valueOf(cs);}
}