Math
Q****9. Palindrome Number
Given an integer x, return trueif x is a palindrome, and false otherwise.
Example 1:
Input: x = 121
Output: true
Explanation: 121 reads as 121 from left to right and from right to left.
Example 2:
Input: x = -121
Output: false
Explanation: From left to right, it reads -121. From right to left, it becomes 121-. Therefore it is not a palindrome.
Example 3:
Input: x = 10
Output: false
Explanation: Reads 01 from right to left. Therefore it is not a palindrome.
Constraints:
-231 <= x <= 231 - 1
Follow up: Could you solve it without converting the integer to a string?
解法及注释:
class Solution { public boolean isPalindrome(int x) { //pre-check if(x < 0 || x > Math.pow(2,31)-1 || (x != 0 && x %10 == 0 )) { return false; } if(x < 10) { return true; } return useString(x); //return useMath(x); } private boolean useString(int x) { String s = String.valueOf(x); int n = s.length(); for(int i = 0; i < n/2; i++) { if(s.charAt(i) != s.charAt(n-i-1)) { return false; } } return true; } private boolean useMath(int x) { int rev = 0; while(x > rev) { rev = rev * 10 + x % 10; x /= 10; } return (x == rev || x == rev/10); }}
Q***66. Plus One
You are given a large integer represented as an integer array digits, where each digits[i] is the ith digit of the integer. The digits are ordered from most significant to least significant in left-to-right order. The large integer does not contain any leading 0's.
Increment the large integer by one and return the resulting array of digits.
Example 1:
Input: digits = [1,2,3]
Output: [1,2,4]
Explanation: The array represents the integer 123.
Incrementing by one gives 123 + 1 = 124.
Thus, the result should be [1,2,4].
Example 2:
Input: digits = [4,3,2,1]
Output: [4,3,2,2]
Explanation: The array represents the integer 4321.
Incrementing by one gives 4321 + 1 = 4322.
Thus, the result should be [4,3,2,2].
Example 3:
Input: digits = [9]
Output: [1,0]
Explanation: The array represents the integer 9.
Incrementing by one gives 9 + 1 = 10.
Thus, the result should be [1,0].
Constraints:
1 <= digits.length <= 1000 <= digits[i] <= 9digitsdoes not contain any leading0's.
解法及注释:
class Solution { public int[] plusOne(int[] digits) { //pre-check if(digits.length == 0 || digits.length > 100) { return digits; } int n = digits.length, curr = 0, add = 0; for(int i = n - 1; i >= 0; i--) { if(add == 1) { curr = digits[i] + add; } else { curr = digits[i] + 1; } if(curr >= 10) { add = 1; digits[i] = 0; } else { add = 0; digits[i] = curr; break; } } if(add != 1) { return digits; } else { int[] res = new int[n + 1]; System.arraycopy(digits, 0, res, 1, n); res[0] = 1; return res; } }}
Q**172. Factorial Trailing Zeroes
Given an integer n, return the number of trailing zeroes inn!.
Note that n! = n * (n - 1) * (n - 2) * ... * 3 * 2 * 1.
Example 1:
Input: n = 3
Output: 0
Explanation: 3! = 6, no trailing zero.
Example 2:
Input: n = 5
Output: 1
Explanation: 5! = 120, one trailing zero.
Example 3:
Input: n = 0
Output: 0
Constraints:
0 <= n <= 104
解法及注释:
class Solution {
public int trailingZeroes(int n) {
//recursion apporach
return n == 0 ? 0 : n/5 + trailingZeroes(n / 5);
//loop approach
//return getTrailingZeros(n);
}
private int getTrailingZeros(int n) {
int count = 0;
while(n > 0) {
n /= 5;
count += n;
}
return count;
}
}
Q***50. Pow(x, n)
Implement pow(x, n), which calculates x raised to the power n (i.e., xn).
Example 1:
Input: x = 2.00000, n = 10
Output: 1024.00000
Example 2:
Input: x = 2.10000, n = 3
Output: 9.26100
Example 3:
Input: x = 2.00000, n = -2
Output: 0.25000
Explanation: 2-2 = 1/22 = 1/4 = 0.25
Constraints:
-100.0 < x < 100.0-231 <= n <= 231-1nis an integer.- Either
xis not zero orn > 0. -104 <= xn <= 104
解法及注释:
class Solution { public double myPow(double x, int n) { int positiveN = Math.abs(n); if(n < 0) return 1 / myPowCalc(x, positiveN); return myPowCalc(x, n); } private double myPowCalc(double x, int n) { if(x == 1 || n == 0) return 1; double temp = myPowCalc(x, n / 2); if(n % 2 == 0) { return temp * temp; } else { return x * temp * temp; } }}
Q***69. Sqrt(x)
Given a non-negative integer x, return the square root of x rounded down to the nearest integer. The returned integer should be non-negative as well.
You must not use any built-in exponent function or operator.
- For example, do not use
pow(x, 0.5)in c++ orx ** 0.5in python.
Example 1:
Input: x = 4
Output: 2
Explanation: The square root of 4 is 2, so we return 2.
Example 2:
Input: x = 8
Output: 2
Explanation: The square root of 8 is 2.82842..., and since we round it down to the nearest integer, 2 is returned.
Constraints:
0 <= x <= 231 - 1
解法及注释:
class Solution { public int mySqrt(int x) { //pre-check if(x <= 0 || x > Math.pow(2,31) - 1) { return 0; } //use binary search int left = 1, right = x; while(left <= right) { int mid = left + (right - left) / 2; if(mid > x / mid) { right = mid - 1; } else if (mid < x / mid) { left = mid + 1; } else { return mid; } } return right; }}
