exercism————Luhn

题目：

解法一：

boolean isVlaid(String s) {
    String number = s.replaceAll(" ","");// check the argumentif (number.length() <= 1 || number.matches("[^0-9]")) {
    throw new IllegalArgumentException("Invalid Argument");}// Converse char to intchar[] set = number.toCharArray();int[] num = new int[set.length];for (int i = 0; i < set.length; i++) {
    num[i] = Character.getNumericValue(set[i]);}// calculate double of every two digitfor (int i = set.length - 2; i >= 0; i -= 2) {
    if (2 * num[i] > 9) {
    num[i] = 2 * num[i] - 9;}else {
    num[i] = 2 * num[i];}}// add all the digitsint sum = 0;for (int i = 0; i < num.length; i++) {
    sum += num[i];}return sum % 10 == 0;}

解法二：

 boolean isValid(String candidate) {
    int[] digits = new StringBuilder(candidate).reverse().toString().replace(" ", "").chars().map(Character::getNumericValue).toArray();if (digits.length < 2){
    return false;}if (Arrays.stream(digits).anyMatch(LuhnValidator::isInvalid)){
    return false;}int sum = IntStream.range(0, digits.length).map(i -> calculateDigitValue(digits[i], i)).sum();return sum % 10 == 0;}private static boolean isInvalid(Integer digit){
    return digit < 0 || digit > 9;}private static Integer calculateDigitValue(Integer digit, Integer index){
    if (index % 2 == 0){
    return digit;}int digitDoubled = digit << 1;if (digitDoubled >= 10){
    return digitDoubled - 9;}return digitDoubled;}

解法三：

public boolean isValid(String str) {
    String s = str.replaceAll(" ", "");String[] nums = s.split("");return s.matches("^\\d{2,}") && getSum(nums) % 10 == 0;}private int getSum(String[] nums) {
    // 如果nums.length为偶数，则从下标0开始，如果nums.length为奇数，则从下标1开始int startDigit = (nums.length & 1) == 0 ? 0 : 1;// 如果为偶数位，则加倍偶数下标；如果位奇数，则加倍奇数下标return IntStream.range(0, nums.length).map(i -> (i & 1) == startDigit ? doubleDigit(Integer.parseInt(nums[i])) : Integer.parseInt(nums[i])).sum();}private int doubleDigit(int x) {
    if (x * 2 > 9) {
    return x * 2 - 9;};return x * 2;}

总结：

• 解法一：中规中矩，正常思路
• 解法二：条理清晰，可读性较强
• 解法三：解法惊奇，找到偶数和奇数长度时，加倍下标也是偶数和奇数，所以只要加倍对应的下标数就ok了，妙！