新增題目2540以及3925解法

Author: clingoram

javascript/LeetCode/Array/2540.js

@@ -0,0 +1,44 @@
+/**
+ * 2540. Minimum Common Value
+ * 
+ * @param {number[]} nums1
+ * @param {number[]} nums2
+ * @return {number}
+ */
+var getCommon = function(nums1, nums2) {
+  /**
+   * 找出兩個陣列中最小的共通元素，找不到則-1
+   * 
+   * 2個陣列長度不一定一樣
+   */
+  // Solution 1.
+  // Set
+  // let set = new Set();
+  // for(const a of nums1){
+  //   set.add(a);
+  // }
+  // for(const b of nums2){
+  //   if(set.has(b)){
+  //     return b;
+  //   }
+  // }
+  // return -1;
+
+
+  // Solution 2.
+  // 2 pointers
+  let i = 0, j = 0;
+  while(i < nums1.length && j < nums2.length){
+    if(nums1[i] < nums2[j]){
+      i++;
+    }else if(nums1[i] > nums2[j]){
+      j++;
+    }else{
+      return nums1[i];
+    }
+  }
+  return -1;
+};
+let nums1 = [1,2,3], nums2 = [2,4];
+// 2
+console.log(getCommon(nums1,nums2));

javascript/LeetCode/Array/3925.js

@@ -0,0 +1,30 @@
+/**
+ * 3925. Concatenate Array With Reverse
+ * 
+ * @param {number[]} nums
+ * @return {number[]}
+ */
+var concatWithReverse = function(nums) {
+    /**
+     * 新陣列長度 = nums.length * 2
+     * 前n個為nums原有排序，後n個為nums反轉後並將它合併一起成一個陣列
+     */
+    let ans = [];
+    for(let i = 0;i < nums.length;++i) {
+      ans.push(nums[i]);
+    }
+    let reverse = nums.reverse();
+    return ans.concat(reverse);
+};
+let nums = [1,2,3];
+/*
+Output: [1,2,3,3,2,1]
+Explanation:
+The first n elements of ans are the same as nums.
+For the next n = 3 elements, each element is taken from nums in reverse order:
+ans[3] = nums[2] = 3
+ans[4] = nums[1] = 2
+ans[5] = nums[0] = 1
+Thus, ans = [1, 2, 3, 3, 2, 1].
+*/
+console.log(concatWithReverse(nums));

javascript/index.js

@@ -360,7 +360,7 @@ var fractionRecurringDecimal = function (a,b) {
 }
 let a = 1,b = 2;
 // "0.5"
-console.log(fractionRecurringDecimal(a,b));
+// console.log(fractionRecurringDecimal(a,b));
 
 /**
  * Recurring Sequence in a Fraction