[khyo] Week 7

longest-substring-without-repeating-characters/higeuni.js

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+/**
+ * @param {string} s
+ * @return {number}
+ * 
+ * complexity
+ * time: O(n)
+ * space: O(n)
+ */
+var lengthOfLongestSubstring = function(s) {
+  let set = new Set();
+  let left = 0;
+  let answer = 0;
+
+  for(let i=0; i<s.length; ++i) {
+      while(set.has(s[i])){
+          set.delete(s[left])
+          if(s[i] === s[left]) {
+              left ++;
+              break;
+          }
+          left ++;
+      }
+      set.add(s[i])
+      answer = Math.max(answer, i - left + 1)
+  }
+  return answer;
+};
+

number-of-islands/higeuni.js

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+/**
+ * @param {character[][]} grid
+ * @return {number}
+ * 
+ * 접근
+ * dfs로 접근
+ * 
+ * 1. 방문한 노드는 0으로 바꿔줘야 함
+ * 2. 방문한 노드는 방문여부를 잘 체크해야 함
+ * 
+ * complexity
+ * time: O(m*n)
+ * space: O(m*n)
+ */
+var numIslands = function(grid) {
+    let answer = 0;
+    const numRows = grid.length;
+    const numCols = grid[0].length;
+
+    const dr = [-1, 1, 0, 0];
+    const dc = [0, 0, -1, 1];
+
+    const dfs = (row, col) => {
+        if (row < 0 || col < 0 || row >= numRows || col >= numCols || grid[row][col] === '0') return;
+
+        grid[row][col] = '0';
+
+        for (let i = 0; i < 4; ++i) {
+            dfs(row + dr[i], col + dc[i]);
+        }
+    };
+
+    for (let r = 0; r < numRows; ++r) {
+        for (let c = 0; c < numCols; ++c) {
+            if (grid[r][c] === '1') {
+                dfs(r, c);
+                answer++;
+            }
+        }
+    }
+
+    return answer;
+};
+

reverse-linked-list/higeuni.js

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+/**
+ * @param {ListNode} head
+ * @return {ListNode}
+ * 
+ * 접근
+ * ListNode 타입에 대해서 공부하고, 링크드 리스트 개념을 접목해서 문제를 풀이
+ * 
+ * complexity
+ * time: O(n)
+ * space: O(1)
+ */
+var reverseList = function(head) {
+  let newList = null
+  let curNode = null
+
+  while(head){
+      curNode = head.next
+      head.next = newList
+      newList = head
+      head = curNode 
+  }
+
+  return newList
+};
+

unique-paths/higeuni.js

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+/**
+ * @param {number} m
+ * @param {number} n
+ * @return {number}
+ * 
+ * 접근 
+ * dynamic programming으로 접근
+ * 초기 점화식 : dp[i][j] = dp[i-1][j] + dp[i][j-1]
+ * 하지만 1차원만 사용해도 되기 때문에 1차원으로 접근
+ * 
+ * complexity
+ * time: O(m*n)
+ * space: O(n)
+ */
+var uniquePaths = function (m, n) {
+  const dp = Array(n).fill(1);
+  for (let i = 1; i < m; ++i) {
+      for (let j = 1; j < n; ++j) {
+          dp[j] += dp[j - 1];
+      }
+  }
+  return dp[n - 1];
+};
+