文章目录
☀️ 前言 ☀️
算法作为极其重要的一点,是大学生毕业找工作的核心竞争力,所以为了不落后与人,开始刷力扣算法题!
🙀 作者简介 🙀
大家好,我是布小禅,一个尽力让无情的代码变得生动有趣的IT小白,很高兴能偶认识你,关注我,每天坚持学点东西,我们以后就是大佬啦!
📢 :❤布小禅❤
📢 作者专栏:
这是我刷第 80/100 道力扣简单题
💗 一、题目描述 💗
将非负整数 num
转换为其对应的英文表示。
示例1:
输入:num = 12345 输出:"Twelve Thousand Three Hundred Forty Five"
示例2:
输入:num = 1234567891 输出:"One Billion Two Hundred Thirty Four Million Five Hundred Sixty Seven Thousand Eight Hundred Ninety One" 来源:力扣(LeetCode) 链接:https://leetcode-cn.com/problems/integer-to-english-words 著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。
提示:0 <= num <= 231 - 1
💁 二、题目解析 💁
思路1
\- 把思路缕清晰 \- 先把英文表示的数字表达出来 \- 然后诸位分析
🏃 三、代码 🏃
☁️ C语言☁️
/* - 把思路缕清晰 - 先把英文表示的数字表达出来 - 然后诸位分析 */ const char *ge[] = { "Zero", "One", "Two", "Three", "Four", "Five", "Six", "Seven", "Eight", "Nine", }; const char *shi1[] = { "Ten", "Eleven", "Twelve", "Thirteen", "Fourteen", "Fifteen", "Sixteen", "Seventeen", "Eighteen", "Nineteen", }; const char *shi2[] = { "Zero", "Ten", "Twenty", "Thirty", "Forty", "Fifty", "Sixty", "Seventy", "Eighty", "Ninety", }; #define N 2000 char result[N] = { 0 }; bool appended = false; void baishige(int n) { int b = (n / 100) % 10; n %= 100; if (b) { // 百位 if (appended) { strcat(result, " "); } strcat(result, ge[b]); strcat(result, " Hundred"); appended = true; } int s = n / 10; n %= 10; if (s) { // 十位 if (appended) { strcat(result, " "); } if (s == 1) { strcat(result, shi1[n]); n = 0; // 不需要再看个位了 } else { strcat(result, shi2[s]); } appended = true; } if (n) { // 个位 if (appended) { strcat(result, " "); } strcat(result, ge[n]); appended = true; } } #define BILLION 1000000000 #define MILLION 1000000 #define THOUSAND 1000 char *numberToWords(int n) { if (n <= 0) { strcpy(result, "Zero"); return result; } appended = false; memset(result, 0, sizeof(result)); int b = n / BILLION; n %= BILLION; if (b) { // BILLION baishige(b); strcat(result, " Billion"); } int m = n / MILLION; n %= MILLION; if (m) { // MILLION baishige(m); strcat(result, " Million"); } int t = n / THOUSAND; n %= THOUSAND; if (t) { // 千位 baishige(t); strcat(result, " Thousand"); } baishige(n); return result; }
🌔 结语 🌔
坚持最重要,每日一题必不可少!😸
期待你的关注和督促!😛