牛客題目連結
1. 題目考點
- 整數溢出判斷
- 符号處理
2. 考點解析
- 分上溢出和下溢出
public int atoi (String str) {
// write code here
if (str == null || str.length() == 0) return 0;
// 處理前導空格
str = str.trim();
// 注意:symbol 預設為 1
int symbol = 1, i = 0;
if (str.charAt(0) == '-') {
symbol = -1;
i = 1;
}
if (str.charAt(0) == '+') i = 1;
int sum = 0, n = 0, max = Integer.MAX_VALUE, min = Integer.MIN_VALUE;
for (; i < str.length(); i++) {
char c = str.charAt(i);
n = symbol * (c - '0');
// 特殊字元處理
if (c > '9') break;
// 上溢出處理
if (sum > max / 10 || (sum == max / 10 && max % 10 < c - '0' ))
return max;
// 下溢出處理
if (sum < min / 10 || (sum == min / 10 && min % 10 > c - '0'))
return min;
sum = sum * 10 + n;
}
return sum;
}
- 統一判斷,
max % 10 = 7,而 min % 10 = 8,當 sum = min,7 < 8 也滿足上溢出條件上溢出包含假下溢出
public int atoi (String str) {
// write code here
if (str == null || str.length() == 0) return 0;
str = str.trim();
int symbol = 1, i = 0;
if (str.charAt(0) == '-') {
symbol = -1;
i = 1;
}
if (str.charAt(0) == '+') i = 1;
int sum = 0, max = Integer.MAX_VALUE, min = Integer.MIN_VALUE;
for (; i < str.length(); i++) {
char c = str.charAt(i);
if (c > '9') break;
if (sum > max / 10 || (sum == max / 10 && max % 10 < c - '0' ))
return symbol == 1 ? max : min;
sum = sum * 10 + c - '0';
}
return symbol * sum;
}