What I Did

LeetCode 알고리즘 문제풀이

Two-Sum

Given an array of integers nums and an integer target, return indices of the two numbers such that they add up to target. You may assume that each input would have exactly one solution, and you may not use the same element twice.

• Find two integers that adds up to target and return their indices.
• Brute Force Way: Straightforward methods of solving a problem trying every possibility rather than advanced techniques to improve efficiency. (this was the first hint)
• I had nothing comes in my mind better than this brute force way to solve this problem. I decided to write a for loop that goes through every elements in the given array(argument) and within that for loop there is another for loop that checks if thoes two numbers add up to target number. If it does, it returns the indices of those two numbers.

public class Two_sum {
    public int[] twoSum(int[] nums, int target) {
        for (int i = 0; i < nums.length; i++) {
            for (int j = i + 1; j < nums.length; j++) {
                if (target - nums[i] == nums[j]) {
                    return new int[]{i, j};
                }
            }
        }
        throw new IllegalArgumentException("No solution");
    }
}

Reverse Integer

Given a signed 32-bit integer x, return x with its digits reversed. If reversing x causes the value to go outside the signed 32-bit integer range [$-2^{31}$, $2^{31} - 1$], then return 0.

• Reverse a given number and return it unless it's out of range [$-2^{31}$, $2^{31} - 1$]
• 먼저 리버스된 값을 담을 reversed라는 long타입 변수로 만든다. int형식으로 받은 인풋도 리버스했을 때 out of range가 될 수 있다. 인풋으로 들어온 x의 10으로 나눴을 때 나머지값을 추가해주는 형식으로 코드를 짰다. 인풋이 음수일 경우를 대비해 sign이라는 변수도 넣어주었다.

public static int reverse(int x) {
        long reversed = 0;
        int sign = 1;

        if (x < 0) { // if x is negative, save that info in sign
            sign *= -1;
            x *= -1;
        }

        while (x > 0) {
            reversed = reversed * 10 + (x % 10);
            x /= 10;
        }

        if (sign == -1) {
            reversed *= sign;
        }

        if ((reversed > Math.pow(-2, 31)) && (reversed < (Math.pow(2, 31) - 1))) {
            return (int) reversed;
        }

        return 0;
    }

Remove Duplicates from Sorted List

Given the head of a sorted linked list, delete all duplicates such that each element appears only once. Return the linked list sorted as well.

• Remove duplicates and sort the list.
• sort 하라고 되어 있는데 테스트 케이스는 non-decresing 이라고 한다. 그럼 sort를 굳이 해줄 필요가 없잖아요,, 조건 왜 붙이셨는지,,
• 헤드부터 시작해 현재 체크중인 노드(current)의 값과 current.next의 값이 같을 시 current.next를 next.next로 설정해준다. (delete)
• 인풋이 비어있을 때를 대비해 current가 null이 되면 조건문을 빠져나오게 해야한다.