[LeetCode 65]Valid Number (通过率最低的一题)

题目链接：valid-number

/** * Validate if a given string is numeric.Some examples:"0" => true" 0.1 " => true"abc" => false"1 a" => false"2e10" => trueNote: It is intended for the problem statement to be ambiguous. You should gather all requirements up front before implementing one. * */public class ValidNumber {//1481 / 1481 test cases passed.//Status: Accepted//Runtime: 214 ms//Submitted: 0 minutes ago//时间复杂度O(n) 空间复杂度O(1)public boolean isNumber(String s) {int i = 0;s = s.trim();// 去除两端的空格boolean hasDecimal = false; // 标记是否已经出现了 小数点(.)boolean hasNum = false; // 标记之前的字符串是否是数值了boolean hasE = false; // 标记是否已经出现了 (e)字符if (s.length() == 0)return false;// 判断是否有正负号字符if (isSign(s.charAt(0))) {i++;}while (i < s.length()) {char c = s.charAt(i);if (isDigit(c)) {if (!hasNum) {hasNum = true;}} else if (isDecimal(c)) {if (hasDecimal || hasE) // 整个数值只能出现一个小数点(.); e的指数不能为小数；return false;hasDecimal = true;} else if (isE(c)) {// 如果之前的字符已经出现(e)字符，，或者(e)出现在整个字符的最前或最后，则都不是一个有效的数值if (hasE || !hasNum || (++i) >= s.length())return false;// 判断指数是否有符号位if (!isSign(s.charAt(i)))i–;hasE = true;hasNum = false; // 需要重新寻找一个数值} elsereturn false; // 如果有 （数字、小数点.、字符e）以外的字符i++;}return hasNum;}public boolean isSign(char c) {return c == '+' || c == '-';}public boolean isDigit(char c) {return (c – '0' >= 0 && c – '0' <= 9);}public boolean isDecimal(char c) {return c == '.';}public boolean isE(char c) {return c == 'e';}public static void main(String[] args) {System.out.println(new ValidNumber().isNumber("0"));System.out.println(new ValidNumber().isNumber(" 0.1 "));System.out.println(new ValidNumber().isNumber("abc"));System.out.println(new ValidNumber().isNumber("1 a"));System.out.println(new ValidNumber().isNumber("2e10"));System.out.println(new ValidNumber().isNumber(".1"));System.out.println(new ValidNumber().isNumber("6e6.5"));}}

痛苦留给的一切，请细加回味!苦难一经过去，苦难就变为甘美。