7. Reverse Integer
Given a 32-bit signed integer, reverse digits of an integer.
Example 1:
Input: 123 Output: 321
提示：原数除以10取余。检查有没有溢出。对于负数，注意C/C++ 采用向零取整的方式，又称为截断取整。
答案:
class Solution {
public:
int reverse(int x) {
int result = 0;
int temp = 0;
while (x)
{
temp = result * 10 + x % 10;
if (result != temp / 10) return 0;
result = temp;
x /= 10;
}
return result;
}
};
8. String to Integer (atoi)
Implement atoi which converts a string to an integer.
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.
Note:
- Only the space character
' 'is considered as whitespace character. - Assume we are dealing with an environment which could only store integers within the 32-bit signed integer range: [−231, 231 − 1]. If the numerical value is out of the range of representable values, INT_MAX (231 − 1) or INT_MIN (−231) is returned.
Example 1:
Input: "42" Output: 42
Example 2:
Input: " -42" Output: -42 Explanation: The first non-whitespace character is '-', which is the minus sign. Then take as many numerical digits as possible, which gets 42.
Example 3:
Input: "4193 with words" Output: 4193 Explanation: Conversion stops at digit '3' as the next character is not a numerical digit.
Example 4:
Input: "words and 987" Output: 0 Explanation: The first non-whitespace character is 'w', which is not a numerical digit or a +/- sign. Therefore no valid conversion could be performed.
Example 5:
Input: "-91283472332" Output: -2147483648 Explanation: The number "-91283472332" is out of the range of a 32-bit signed integer. Thefore INT_MIN (−231) is returned.提示：
答案:
9. Palindrome Number
Determine whether an integer is a palindrome. An integer is a palindrome when it reads the same backward as forward.
Example 1:
Input: 121 Output: true
提示：方法一：res保存反转之后的值，对比x与res,如果反转溢出那么其也不相等
方法二：用res保存x的一半，比较res与x（注意奇偶数的情况）
答案:
class Solution {
public:
bool isPalindrome(int x) {
if(x < 0) return false;
int y = x;
int res = 0;
while(y != 0) {
res = res * 10 + y % 10;
y /= 10;
}
return x == res;
}
};
class Solution {
public:
bool isPalindrome(int x) {
if (x<0 || (x!=0 && x%10==0)) return false;
int rev = 0;
while (x>rev){
rev = rev*10 + x%10;
x = x/10;
}
return (x==rev || x==rev/10);
}
};
12. Integer to Roman
Roman numerals are represented by seven different symbols: I, V, X, L, C, D and M.
Symbol Value I 1 V 5 X 10 L 50 C 100 D 500 M 1000
For example, two is written as II in Roman numeral, just two one's added together. Twelve is written as, XII, which is simply X + II. The number twenty seven is written as XXVII, which is XX + V + II.
Roman numerals are usually written largest to smallest from left to right. However, the numeral for four is not IIII. Instead, the number four is written as IV. Because the one is before the five we subtract it making four. The same principle applies to the number nine, which is written as IX. There are six instances where subtraction is used:
Ican be placed beforeV(5) andX(10) to make 4 and 9.Xcan be placed beforeL(50) andC(100) to make 40 and 90.Ccan be placed beforeD(500) andM(1000) to make 400 and 900.
Given an integer, convert it to a roman numeral. Input is guaranteed to be within the range from 1 to 3999.
Example 4:
Input: 58 Output: "LVIII" Explanation: C = 100, L = 50, XXX = 30 and III = 3.
Example 5:
Input: 1994 Output: "MCMXCIV" Explanation: M = 1000, CM = 900, XC = 90 and IV = 4.
提示：字符串数组保存所有的情况
答案:
class Solution {
public:
string intToRoman(int num) {
static const string s[4][10]=
{
{"","I","II","III","IV","V","VI","VII","VIII","IX"},
{"","X","XX","XXX","XL","L","LX","LXX","LXXX","XC"},
{"","C","CC","CCC","CD","D","DC","DCC","DCCC","CM"},
{"","M","MM","MMM"},
};
return s[3][num/1000] + s[2][num/100%10] + s[1][num/10%10] + s[0][num%10];
}
};
13. Roman to Integer
Roman numerals are represented by seven different symbols: I, V, X, L, C, D and M.
Symbol Value I 1 V 5 X 10 L 50 C 100 D 500 M 1000
提示：用unordered_map<char, int>或者roman['X' - 'A']保存所有情况，当前数比后面的数小时，要减去当前数
答案:
class Solution {
public:
int romanToInt(string s)
{
if(s.empty()) return 0;
unordered_map<char, int> T = { { 'I' , 1 },
{ 'V' , 5 },
{ 'X' , 10 },
{ 'L' , 50 },
{ 'C' , 100 },
{ 'D' , 500 },
{ 'M' , 1000 } };
int sum = T[s.back()];
for (int i = s.length() - 2; i >= 0; --i)
{
if (T[s[i]] < T[s[i + 1]])
{
sum -= T[s[i]];
}
else
{
sum += T[s[i]];
}
}
return sum;
}
};
29. Divide Two Integers
Given two integers dividend and divisor, divide two integers without using multiplication, division and mod operator.
Return the quotient after dividing dividend by divisor.
The integer division should truncate toward zero.
Example 1:
Input: dividend = 10, divisor = 3 Output: 3
提示：溢出的情况：除数为0；被除数为INT_MIN,除数为-1。
商的符号由两者与零比较的异或，商的大小为通过1左移为操作求得。
答案:
class Solution {
public:
int divide(int dividend, int divisor) {
if (!divisor || (dividend == INT_MIN && divisor == -1))
return INT_MAX;
int sign = ((dividend < 0) ^ (divisor < 0)) ? -1 : 1;
long long dvd = labs(dividend);
long long dvs = labs(divisor);
int res = 0;
while (dvd >= dvs) {
long long temp = dvs, multiple = 1;
while (dvd >= (temp << 1)) {
temp <<= 1;
multiple <<= 1;
}
dvd -= temp;
res += multiple;
}
return sign == 1 ? res : -res;
}
};
43. Multiply Strings
Given two non-negative integers num1 and num2 represented as strings, return the product of num1 and num2, also represented as a string.
Example 1:
Input: num1 = "2", num2 = "3" Output: "6"
Example 2:
Input: num1 = "123", num2 = "456" Output: "56088"
Note:
- The length of both
num1andnum2is < 110. - Both
num1andnum2contain only digits0-9. - Both
num1andnum2do not contain any leading zero, except the number 0 itself. - You must not use any built-in BigInteger library or convert the inputs to integer directly.
提示：
(sum[i + j + 1] - '0') + (num1[i] - '0') * (num2[j] - '0') + carry;
排除单个0的情况。
答案:
class Solution {
public:
string multiply(string num1, string num2) {
string sum(num1.size() + num2.size(), '0');
for (int i = num1.size() - 1; 0 <= i; --i) {
int carry = 0;
for (int j = num2.size() - 1; 0 <= j; --j) {
int tmp = (sum[i + j + 1] - '0') + (num1[i] - '0') * (num2[j] - '0') + carry;
sum[i + j + 1] = tmp % 10 + '0';
carry = tmp / 10;
}
sum[i] += carry;
}
size_t startpos = sum.find_first_not_of("0");
if (string::npos != startpos) {
return sum.substr(startpos);
}
return "0";
}
};
50. Pow(x, n)
Implement pow(x, n), which calculates x raised to the power n (xn).
Example 1:
Input: 2.00000, 10 Output: 1024.00000
Example 2:
Input: 2.10000, 3 Output: 9.26100
Example 3:
Input: 2.00000, -2 Output: 0.25000 Explanation: 2-2 = 1/22 = 1/4 = 0.25
Note:
- -100.0 < x < 100.0
- n is a 32-bit signed integer, within the range [−231, 231 − 1]
提示：当n ==0,返回1,这也是递归的出口。考虑n的奇偶与正负。
答案:
class Solution {
public:
double myPow(double x, int n) {
if (n==0) return 1;
double t = myPow(x,n/2);//整除很关键
if (n%2) {
return n<0 ? 1/x*t*t : x*t*t;
} else {
return t*t;
}
}
};
60. Permutation Sequence
The set [1,2,3,...,n] contains a total of n! unique permutations.
By listing and labeling all of the permutations in order, we get the following sequence for n = 3:
"123""132""213""231""312""321"
Given n and k, return the kth permutation sequence.
Note:
- Given n will be between 1 and 9 inclusive.
- Given k will be between 1 and n! inclusive.
提示：
答案：