update solution

Author: Neil94n

solution/0001.Two Sum/Solution3.js

@@ -0,0 +1,9 @@
+var twoSum = function(nums, target) {
+    const map = new Map()
+    for(let i = 0; i < nums.length; i++){
+        if (map.has(target - nums[i])){
+            return [ map.get(target - nums[i]), i ]
+        }
+        map.set(nums[i], i)
+    }
+};

solution/0002.Add Two Numbers/Solution2.js

@@ -0,0 +1,46 @@
+/**
+ * Definition for singly-linked list.
+ * function ListNode(val) {
+ *     this.val = val;
+ *     this.next = null;
+ * }
+ */
+/**
+ * @param {ListNode} l1
+ * @param {ListNode} l2
+ * @return {ListNode}
+ */
+var addTwoNumbers = function(l1, l2) {
+    let head = new ListNode(0)
+       let cur = head
+       let curry = 0
+   
+       while (true) {
+           let sum = curry
+           sum += l1 ? l1.val : 0
+           sum += l2 ? l2.val : 0
+           cur.val = sum % 10
+           curry = parseInt(sum / 10)
+           if (l1) l1 = l1.next
+           if (l2) l2 = l2.next
+           if (l1 != null || l2 != null) {
+               cur.next = new ListNode(0)
+               cur = cur.next
+           } else {
+               break
+           }
+       }
+       if (curry != 0) {
+           cur.next = new ListNode(0)
+           cur = cur.next
+           cur.val = curry
+       }
+       return head
+   };
+   
+   var l1 = new ListNode(1)
+   l1.next = new ListNode(8)
+   
+   var l2 = new ListNode(0)
+   
+   console.log(addTwoNumbers(l1, l2))

solution/0003.Longest Substring Without Repeating Characters/Solution.js

@@ -0,0 +1,22 @@
+/**
+ * @param {string} s
+ * @return {number}
+ */
+var lengthOfLongestSubstring = function(s) {
+    var start = 0; // 非重复字符串开始索引
+    var max = 0; // 最长字符串长度
+    var visitedCharByPosition = {};
+    for (var position = 0; position < s.length; position++) {
+        var nextChar = s[position];
+        if (nextChar in visitedCharByPosition && visitedCharByPosition[nextChar] >= start) {
+            // 有重复，非重复字符串索引从下一个 index 开始
+            start = visitedCharByPosition[nextChar] + 1;
+            visitedCharByPosition[nextChar] = position;
+        } else {
+            visitedCharByPosition[nextChar] = position;
+            // 非重复，求非重复值
+            max = Math.max(max, position + 1 - start);
+        }
+    }
+    return max;
+};