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Implement atoi to convert a string to an integer.
Hint: Carefully consider all possible input cases. If you want a challenge, please do not see below and ask yourself what are the possible input cases.
Notes: It is intended for this problem to be specified vaguely (ie, no given input specs). You are responsible to gather all the input requirements up front.
正如题目所说,真的是考虑各种情况。
public int myAtoi(String str) { if (str == null || str.length() == 0) return 0; str = str.trim(); if (str.matches("([0-9]|-|\\+)[0-9]*")) { if (str.charAt(0) == '+') { if (str.length() > 11 || str.length() == 1 || (str.length() == 11 && !str .matches("[+][1-2][0-1][0-4][0-7][0-4][0-8][0-3][0-6][0-4][0-7]"))) return 0; else return Integer.parseInt(str); } if (str.charAt(0) == '-') { if (str.length() > 11 || str.length() == 1 || (str.length() == 11 && !str .matches("[-][1-2][0-1][0-4][0-7][0-4][0-8][0-3][0-6][0-4][0-8]"))) { return 0; } else { return Integer.parseInt(str); } } if ((str.charAt(0) + "").matches("[0-9]")) { if (str.length() > 10 || (str.length() == 10 && !str .matches("[1-2][0-1][0-4][0-7][0-4][0-8][0-3][0-6][0-4][0-7]"))) return 0; else return Integer.parseInt(str); } } return 0; }
上述程序是我第一次想出来的,基本情况都考虑进去了,还差一种情况没有考虑到:
下面这个就考虑到了所有情况,而且效率高很多。
public int myAtoi(String str) { if (str == null) return 0; str = str.trim(); if (str.length() == 0) return 0; int index = 0; int sign = 1; if (str.charAt(index) == '+') { sign = -1; index++; } else if (str.charAt(index) == '-') index++; long num = 0; for (; index < str.length(); index++) { System.out.println(str.charAt(index)); if (str.charAt(index) < '0' || str.charAt(index) > '9') break; num = num * 10 + (str.charAt(index) - '0'); if (num > Integer.MAX_VALUE) break; } if (num * sign >= Integer.MAX_VALUE) return Integer.MAX_VALUE; if (num * sign <= Integer.MIN_VALUE) return Integer.MIN_VALUE; return (int) num * sign; }