Emp*_*mp1 7 c++ arrays sorting string algorithm
Run Code Online (Sandbox Code Playgroud)String str[]={"-123","89","-10","456"};
str是一个字符串数组,每个字符串都是整数格式,您必须及时对该数组进行排序O(n log n)。中的字符串
str可以表示正整数和负整数。这些字符串的最大长度为1024个字符。
我知道解决此问题的一种方法是将字符串转换为数字,然后将它们进行比较。还有其他解决方案吗?
use*_*109 12
另一个解决方案是实现自己的比较功能:
-的字符串-
是较小的数字。-,则比较字符串的长度。字符串越长,数字越小。如果两个字符串的长度相同,请执行标准字符串比较,但取反结果。dru*_*nly 11
这是一个最小且可能不足(无法处理前导零,空白等)的示例,可以满足您的要求。
这些评论说明了它在做什么。:)
#include <algorithm>
#include <iostream>
#include <string>
#include <vector>
int main() {
std::vector<std::string> strings = {
"-1", "-1", "-20", "-4", "3", "0", "-0", "1", "20", "20", "44020",
};
// Assumes everything in "strings" has no whitespace in it.
// Assumes everything in "strings" does not have leading zeroes.
// Assumes everything in "strings" is an ascii representaion of an integer.
// Assumes everything in "strings" is nonempty.
std::sort(strings.begin(), strings.end(),
[](const std::string &a, const std::string &b) {
const bool a_is_negative = a[0] == '-';
const bool b_is_negative = b[0] == '-';
if (a_is_negative != b_is_negative) {
// If they have different signs, then whichever is negative is
// smaller.
return a_is_negative;
} else if (a.length() != b.length()) {
// If they have the same sign, then whichever has more
// characters is larger in magnitude. When the sign is negative,
// the longer (more digits) number is "more negative". When
// positive, the longer (more digits) number is "more positive".
return (a.length() < b.length()) != a_is_negative;
} else {
// Otherwise a lexicographic comparison of "a" and "b" will
// determine which string is larger in magnitude. Using the same
// logic above, we account for the "negative vs. positive"
// comparison.
return (a < b) != a_is_negative;
}
});
for (const auto &str : strings) {
std::cout << str << " ";
}
std::cout << std::endl;
}
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