首页 > 代码库 > reconstruct-original-digits-from-english(好)
reconstruct-original-digits-from-english(好)
https://leetcode.com/problems/reconstruct-original-digits-from-english/ //https://discuss.leetcode.com/topic/63386/one-pass-o-n-java-solution-simple-and-clear public class Solution { // zero one two three four five six seven eight nine ten // z 0 // e 0 1 3 3 5 7 7 8 9 // r 0 3 4 // o 0 1 2 4 // n 1 7 9 9 // t 2 3 8 // w 2 // h 3 8 // f 4 5 // u 4 // i 5 6 8 9 // v 5 7 // s 6 7 // x 6 // g 8 public String originalDigits(String s) { // 太牛了,开始我也想到统计,但是想到删除字符串那些复杂的操作上面去了 int[] counts = new int[10]; for (int i=0; i<s.length(); i++) { char ch = s.charAt(i); if (ch == ‘z‘) counts[0]++; if (ch == ‘u‘) counts[4]++; if (ch == ‘w‘) counts[2]++; if (ch == ‘x‘) counts[6]++; if (ch == ‘g‘) counts[8]++; if (ch == ‘h‘) counts[3]++; // 3, 8 if (ch == ‘s‘) counts[7]++; // 6, 7 if (ch == ‘f‘) counts[5]++; // 4, 5 if (ch == ‘o‘) counts[1]++; // 0, 1, 2, 4 if (ch == ‘i‘) counts[9]++; // 5, 6, 8, 9 } counts[3] -= counts[8]; counts[7] -= counts[6]; counts[5] -= counts[4]; counts[1] -= counts[0] + counts[2] + counts[4]; counts[9] -= counts[5] + counts[6] + counts[8]; StringBuilder sb = new StringBuilder(); for (int i=0; i<10; i++) { for (int j=0; j<counts[i]; j++) { sb.append(i); } } return sb.toString(); } }
reconstruct-original-digits-from-english(好)
声明:以上内容来自用户投稿及互联网公开渠道收集整理发布,本网站不拥有所有权,未作人工编辑处理,也不承担相关法律责任,若内容有误或涉及侵权可进行投诉: 投诉/举报 工作人员会在5个工作日内联系你,一经查实,本站将立刻删除涉嫌侵权内容。