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23 changes: 23 additions & 0 deletions combination-sum/s0ooo0k.java
Original file line number Diff line number Diff line change
@@ -0,0 +1,23 @@
class Solution {
List<List<Integer>> result = new ArrayList<>();
int[] candidates;

public List<List<Integer>> combinationSum(int[] candidates, int target) {
this.candidates=candidates;
backtrack(new ArrayList<>(), target, 0);
return result;
}
void backtrack(List<Integer> comb, int remain, int start) {
if(remain==0) {
result.add(new ArrayList<>(comb));
return;
}
if(remain<0) return;
for(int i=start; i<candidates.length; i++) {
comb.add(candidates[i]);
backtrack(comb, remain-candidates[i], i);
comb.remove(comb.size() - 1);
}
}
}

22 changes: 22 additions & 0 deletions decode-ways/s0ooo0k.java
Original file line number Diff line number Diff line change
@@ -0,0 +1,22 @@
class Solution {
public int numDecodings(String s) {
if(s.charAt(0)=='0') return 0;

int n = s.length();
int[] dp = new int[n+1];

dp[0]=1;
dp[1]=1;

for(int i=2; i<=n; i++) {
if(s.charAt(i-1)!='0') dp[i]+=dp[i-1];

int group = Integer.parseInt(s.substring(i-2, i));
if(group>=10 && group<=26) {
dp[i]+=dp[i-2];
}
}
return dp[n];
}
}

13 changes: 13 additions & 0 deletions maximum-subarray/s0ooo0k.java
Original file line number Diff line number Diff line change
@@ -0,0 +1,13 @@
class Solution {
public int maxSubArray(int[] nums) {
int curr = nums[0];
int maxSum = nums[0];

for(int i=1; i<nums.length; i++) {
curr = Math.max(nums[i], curr+nums[i]);
maxSum = Math.max(maxSum, curr);
}
return maxSum;
}
}

14 changes: 14 additions & 0 deletions number-of-1-bits/s0ooo0k.java
Original file line number Diff line number Diff line change
@@ -0,0 +1,14 @@
class Solution {
public int hammingWeight(int n) {
/*
* 시간복잡도 O(log n)
*/
int cnt = 0;
while(n!=0) {
if(n%2==1) cnt++;
n = n / 2;
}
return cnt;
}
}

30 changes: 30 additions & 0 deletions valid-palindrome/s0ooo0k.java
Original file line number Diff line number Diff line change
@@ -0,0 +1,30 @@
class Solution {
/*
* 초반 풀이 시 while문 내에서 left<right 미체크로 오류 / 조건 추가
* 중복 while문 사용하지 않는 방법이 있는지 고민
*
* 시간복잡도 : O(n)
*/
public boolean isPalindrome(String s) {
int left = 0;
int right = s.length()-1;

while(left < right) {
while(left<right && !Character.isLetterOrDigit(s.charAt(left))) {
left++;
}
while(left<right && !Character.isLetterOrDigit(s.charAt(right))) {
right--;
}

if (Character.toLowerCase(s.charAt(left)) != Character.toLowerCase(s.charAt(right))) {
return false;
}

left++;
right--;
}
return true;
}
}