[영우] Week7 Solutions

longest-substring-without-repeating-characters/yeonguchoe.cpp

@@ -0,0 +1,27 @@
+class Solution {
+public:
+    int lengthOfLongestSubstring(string s) {
+
+        int current_max = 0;
+        int pivot = 0;
+        int checker = 0;
+
+        int current_length = 0;
+        while (pivot < s.size()) {
+            set<char> included;
+            included.insert(s[pivot]);
+            checker = pivot + 1;
+            while (checker < s.size() and
+                   included.find(s[checker]) == included.end()) {
+                included.insert(s[checker]);
+                checker += 1;
+            }
+            current_length = checker - pivot;
+            current_max = max(current_max, current_length);
+            pivot += 1;
+        }
+        return current_max;
+    }
+    // 시간 복잡도: O(n)
+    // 공간 복잡도: O(n)
+};

number-of-islands/yeonguchoe.cpp

@@ -0,0 +1,68 @@
+class Solution {
+public:
+    int numIslands(vector<vector<char>>& grid) {
+        int count = 0;
+
+        int row_size = grid.size();
+        int column_size = grid[0].size();
+
+        queue<pair<int, int>> q;
+
+        for (int i = 0; i < row_size; i++) {
+            for (int j = 0; j < column_size; j++) {
+                if (grid[i][j] == '1') {
+                    LookAround(i, j, grid, q, row_size, column_size);
+                    count += 1;
+                }
+            }
+        }
+        return count;
+    }
+
+    // BFS
+    void LookAround(int& x, int& y, vector<vector<char>>& grid,
+                    queue<pair<int, int>>& q, int& row_size, int& column_size) {
+        pair<int, int> center_location{x, y};
+        q.push(center_location);
+
+        grid[x][y] = '0';
+
+        while (!q.empty()) {
+            pair<int, int> selected_location = q.front();
+            q.pop();
+
+            int selected_x = selected_location.first;
+            int selected_y = selected_location.second;
+
+            // 선택된 cell에서 왼쪽
+            if (selected_x - 1 >= 0 &&
+                grid[selected_x - 1][selected_y] == '1') {
+                q.push({selected_x - 1, selected_y});
+                grid[selected_x - 1][selected_y] = '0';
+            }
+
+            // 선택된 cell에서 오른쪽
+            if (selected_x + 1 < row_size &&
+                grid[selected_x + 1][selected_y] == '1') {
+                q.push({selected_x + 1, selected_y});
+                grid[selected_x + 1][selected_y] = '0';
+            }
+
+            // 선택된 cell에서 위쪽
+            if (selected_y - 1 >= 0 &&
+                grid[selected_x][selected_y - 1] == '1') {
+                q.push({selected_x, selected_y - 1});
+                grid[selected_x][selected_y - 1] = '0';
+            }
+
+            // 선택된 cell에서 아래쪽
+            if (selected_y + 1 < column_size &&
+                grid[selected_x][selected_y + 1] == '1') {
+                q.push({selected_x, selected_y + 1});
+                grid[selected_x][selected_y + 1] = '0';
+            }
+        }
+    }
+    // 시간 복잡도: O(row*column)
+    // 공간 복잡도: O(row*column)
+};

reverse-linked-list/yeonguchoe.c

@@ -0,0 +1,25 @@
+/**
+ * Definition for singly-linked list.
+ * struct ListNode {
+ *     int val;
+ *     struct ListNode *next;
+ * };
+ */
+struct ListNode* reverseList(struct ListNode* head) {
+
+    struct ListNode* ptr1 = NULL;
+    struct ListNode* ptr2 = head;
+    struct ListNode* ptr3 = NULL;
+
+    while (ptr2 != NULL) {
+        ptr3 = ptr2->next;
+        ptr2->next = ptr1;
+
+        // 앞으로 전진
+        ptr1 = ptr2;
+        ptr2 = ptr3;
+    }
+    return ptr1;
+}
+// 시간 복잡도: O(노드 개수)
+// 공간 복잡도: O(1)

set-matrix-zeroes/yeonguchoe.c

@@ -0,0 +1,53 @@
+void setZeroes(int** matrix, int matrixSize, int* matrixColSize) {
+
+    int row_size = matrixSize;
+    int column_size = matrixColSize[0];
+
+    bool zero_on_first_row = false;
+    bool zero_on_first_column = false;
+
+    // 첫번째 row에 0이 존재하는지 확인
+    for (int i = 0; i < column_size; i++) {
+        if (matrix[0][i] == 0) {
+            zero_on_first_row = true;
+        }
+    }
+    // 첫번째 column에 0이 존재하는지 확인
+    for (int i = 0; i < row_size; i++) {
+        if (matrix[i][0] == 0) {
+            zero_on_first_column = true;
+        }
+    }
+
+    // 전체 돌면서 0 발견시 첫번째 row, column에 0으로 표시
+    for (int i = 0; i < row_size; i++) {
+        for (int j = 0; j < column_size; j++) {
+            if (matrix[i][j] == 0) {
+                matrix[0][j] = 0;
+                matrix[i][0] = 0;
+            }
+        }
+    }
+
+    for (int i = 1; i < row_size; i++) {
+        for (int j = 1; j < column_size; j++) {
+            if (matrix[0][j] == 0 || matrix[i][0] == 0) {
+                matrix[i][j] = 0;
+            }
+        }
+    }
+
+    if (zero_on_first_row) {
+        for (int i = 0; i < column_size; i++) {
+            matrix[0][i] = 0;
+        }
+    }
+
+    if (zero_on_first_column) {
+        for (int i = 0; i < row_size; i++) {
+            matrix[i][0] = 0;
+        }
+    }
+}
+// 시간 복잡도: O(matrix 크기)
+// 공간 복잡도: O(1)

unique-paths/yeonguchoe.cpp

@@ -0,0 +1,14 @@
+class Solution {
+public:
+    int combination(int n, int r) {
+        int k = n - r > r ? r : n - r;
+        unsigned long long result = 1;
+        for (int i = 0; i < k; i++) {
+            result = result * (n - i) / (i + 1);
+        }
+        return result;
+    }
+    int uniquePaths(int m, int n) { return combination(m - 1 + n - 1, m - 1); }
+    // 시간 복잡도: O(min(m,n))
+    // 공간 복잡도: O(1)
+};