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35 changes: 35 additions & 0 deletions 3sum/mintheon.java
Original file line number Diff line number Diff line change
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import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashSet;
import java.util.List;
import java.util.Set;

class Solution {
public List<List<Integer>> threeSum(int[] nums) {
Arrays.sort(nums);

Set<List<Integer>> answer = new HashSet<>();

for(int i = 0; i < nums.length - 2; i++) {
int left = i + 1;
int right = nums.length - 1;

while(left < right) {
int sum = nums[i] + nums[left] + nums[right];

if(sum < 0) {
left++;
} else if(sum > 0) {
right--;
} else{
answer.add(List.of(nums[i], nums[left], nums[right]));

left++;
right--;
}
}
}

return new ArrayList<>(answer);
}
}
17 changes: 17 additions & 0 deletions climbing-stairs/mintheon.java
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class Solution {
public int climbStairs(int n) {
int[] stepCount = new int[n + 1];

if(n == 1) {
return 1;
}
Comment on lines +5 to +7
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초기값을 이용한 early return인데, 저라면 dp를 잘 이해하고 있다고 어필하기 위해 n == 2 인 경우도 추가할 것 같습니다.


stepCount[1] = 1;
stepCount[2] = 2;
for(int i = 3; i <= n; i++) {
stepCount[i] = stepCount[i - 1] + stepCount[i - 2];
}

return stepCount[n];
}
}
13 changes: 13 additions & 0 deletions valid-anagram/mintheon.java
Original file line number Diff line number Diff line change
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import java.util.Arrays;

class Solution {
public boolean isAnagram(String s, String t) {
char[] sChar = s.toCharArray();
char[] tChar = t.toCharArray();

Arrays.sort(sChar);
Arrays.sort(tChar);

return Arrays.equals(sChar, tChar);
}
}
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