Leetcode -- Day 45

Reverse num/word/bits

最近回国了 有点懒

Question 1

Reverse Integer

Reverse digits of an integer.

Example1: x = 123, return 321
Example2: x = -123, return -321

Be careful about some interesting cases. i.e.negative number, overflow case.

    public int reverse(int x) {
        if (x >= -9 && x <= 9){
            return x;
        }
        
        boolean isNeg = false;
        if (x < 0){
            isNeg = true;
            x = 0 - x;
        }
        
        double result = 0;
        
        while(x > 0){
            result = result * 10 + x % 10;
            x = x / 10;
        }
        
        if (isNeg){
            return -result < Integer.MIN_VALUE ? 0 : -(int)result;
        }
        else{
            return result > Integer.MAX_VALUE ? 0 : (int)result;
        }
    } 

Question 2

Reverse Words in a String

Given an input string, reverse the string word by word.

For example,
Given s = "the sky is blue",
return "blue is sky the".

I use split to change it an word array, or you can use toCharArray() to change it to char array. 

    public String reverseWords(String s) {
        StringBuffer result = "";
        String[] array = s.split(" ");
        for (int i = array.length - 1; i >= 0; i --){
            String temp = array[i].trim();
            if (temp.equals("") == false){
                if (i != array.length - 1){
                    result += " " + temp;
                }
                else{
                    result += temp;
                }
            }
        }
        return result.trim();
    }

 

Question 3

Reverse Words in a String II
这里要求in-place，也就是说不需要开辟额外空间

Reverse all and then deal with space.

public class Solution {
    public void reverseWords(char[] s) {
        reverse(s, 0, s.length);//reverse the whole char array
        //i is the start reverse point, j is the end reverse pont, i's change is
        //controlled by j, when j is the space, set i to the character taht next to space, i.e. i+1
        for (int i=0, j=0; j<=s.length; j++) {
            if (j==s.length || s[j]==' ') {
                reverse(s, i, j);
                i =  j + 1; //j is the space, specify i to j+1, i.e. the character next to space
            }
        }
    }
    //begin: indicate the start reversing location
    //end-1:last character, end-1-i: i is the distance to end
    private void reverse(char [] s, int begin, int end) {
        //[!]i<(end-begin)/2
        for (int i=0; i<(end-begin)/2; i++) {
            char temp = s[begin+i];
            s[begin+i] = s[end-i-1];
            s[end-i-1] = temp;
        }
    }
}

Question 4

Reverse Bits

Reverse bits of a given 32 bits unsigned integer.

For example, given input 43261596 (represented in binary as 00000010100101000001111010011100), return 964176192 (represented in binary as00111001011110000010100101000000).

n & 1 means always get the first digit of n. And put it to reverse's first digit (reversed << 1 menas move all digits to left by 1 bit). 

    public int reverseBits(int n) {
        int reversed = 0;
        for (int i = 0; i < 32; i ++){
           reversed = (reversed << 1) | (n & 1);
            n = n >> 1; 
        }
        return reversed;
    }