資訊專欄INFORMATION COLUMN
摘要：難度是標(biāo)準(zhǔn)庫中的一個(gè)函數(shù)可以將字符串表示的整數(shù)轉(zhuǎn)換為現(xiàn)在要求我們自己來實(shí)現(xiàn)它解題過程中主要有以下兩點(diǎn)需要注意字符串開頭可能出現(xiàn)或者需要處理使用來記錄中間結(jié)果防止溢出下面是的解法
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.Requirements for atoi: The function first discards as many whitespace
characters as necessary until the first non-whitespace character is found. Then, starting from this character, takes an optional initial plus or minus sign followed by as many numerical digits as possible,
and interprets them as a numerical value.The string can contain additional characters after those that form the
integral number, which are ignored and have no effect on the behavior
of this function.If the first sequence of non-whitespace characters in str is not a
valid integral number, or if no such sequence exists because either
str is empty or it contains only whitespace characters, no conversion
is performed.If no valid conversion could be performed, a zero value is returned.
If the correct value is out of the range of representable values,
INT_MAX (2147483647) or INT_MIN (-2147483648) is returned.
難度: Easy
atoi是c標(biāo)準(zhǔn)庫中的一個(gè)函數(shù), 可以將字符串表示的整數(shù)轉(zhuǎn)換為int. 現(xiàn)在要求我們自己來實(shí)現(xiàn)它.
解題過程中, 主要有以下兩點(diǎn)需要注意:
字符串開頭可能出現(xiàn) + 或者 - , 需要處理
使用long來記錄中間結(jié)果防止溢出
下面是AC的解法:
public class Solution {
public int myAtoi(String str) {
str = str.trim();
if (str.length() == 0) {
return 0;
}
int flag = 1;
int start = 0;
long sum = 0;
if (str.charAt(0) == "-") {
flag = -1;
start = 1;
} else if (str.charAt(0) == "+") {
flag = 1;
start = 1;
}
for (int i = start; i < str.length(); i++) {
if (str.charAt(i) > "9" || str.charAt(i) < "0") {
return this.getSuitInt(sum * flag);
}
sum = sum * 10 + (str.charAt(i) - "0");
if (sum >= Integer.MAX_VALUE) {
break;
}
}
return this.getSuitInt(sum * flag);
}
private int getSuitInt(long sum) {
if (sum > Integer.MAX_VALUE) {
return Integer.MAX_VALUE;
} else if (sum < Integer.MIN_VALUE) {
return Integer.MIN_VALUE;
} else {
return (int) sum;
}
}
public static void main(String[] args) {
Solution s = new Solution();
System.out.println(s.myAtoi("0"));
System.out.println(s.myAtoi("4"));
System.out.println(s.myAtoi("18384"));
System.out.println(s.myAtoi("-10203"));
System.out.println(s.myAtoi("+304040"));
System.out.println(s.myAtoi("9223372036854775809"));
}
}
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