practice to slove a problem

Author: clingoram

javascript/index.js

@@ -1061,21 +1061,27 @@ let numerator = 1, denominator = 2;
  * @return {number[]}
  */
 var findXSum = function(nums, k, x) {
-	// 將nums切成長度 = k的，並計算在裡面有幾個是連續子陣列且元素出現次數 = x的
+	// 將nums切成長度 = k的陣列，並計算在該陣列裡面有幾個元素出現次數 = x的
 	// 最後將符合的加總，並組成新陣列，形成元素i
-    let res = [];
-	let map = new Map();
-
-	for(let i = 0;i < k;++i) {
-		map.has(nums[i]) ? map.set(nums[i],map.get(nums[i]) + 1) : map.set(nums[i],1);
-	}
+  let res = [];
+	// let map = new Map();
+	// for(let i = 0;i < k;++i) {
+	// 	map.has(nums[i]) ? map.set(nums[i],map.get(nums[i]) + 1) : map.set(nums[i],1);
+	// }
+  let i = 0;
+  while(i < k){
+    let sub = nums[i];
+    console.log(sub);
+    i++;
+  }
+  // for(let i = 0;i < k;++i) {
+  //   console.log(nums[i])
+  // }
 
 };
-// let nums = [1,1,2,2,3,4,2,3], k = 6, x = 2;
+let nums = [1,1,2,2,3,4,2,3], k = 6, x = 2;
 // [6,10,12]
 // For subarray [1, 1, 2, 2, 3, 4], only elements 1 and 2 will be kept in the resulting array. Hence, answer[0] = 1 + 1 + 2 + 2.
 // For subarray [1, 2, 2, 3, 4, 2], only elements 2 and 4 will be kept in the resulting array. Hence, answer[1] = 2 + 2 + 2 + 4. Note that 4 is kept in the array since it is bigger than 3 and 1 which occur the same number of times.
 // For subarray [2, 2, 3, 4, 2, 3], only elements 2 and 3 are kept in the resulting array. Hence, answer[2] = 2 + 2 + 2 + 3 + 3.
-// console.log(findXSum(nums,k,x));
-
-
+console.log(findXSum(nums,k,x));