Add solution 0905 solution.js

solution/0905.Sort Array By Parity/Solution.js

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+/**
+ * @param {number[]} A
+ * @return {number[]}
+ */
+// 第一次的做法
+var sortArrayByParity = function (A) {
+  const len = A.length;
+  if (len == 1) return A;
+  let evenNumber = [];
+  let oddNumber = [];
+  for (let i = 0; i < len; i++) {
+    if (A[i] % 2 == 0) {
+      evenNumber.push(A[i]);
+    }
+    if (A[i] % 2 != 0) {
+      oddNumber.push(A[i]);
+    }
+  }
+  return evenNumber.concat(oddNumber);
+};
+// 修改第一次的代码,只使用一个数组，减少一次合并数组操作
+var sortArrayByParity = function (A) {
+  const len = A.length;
+  if (len == 1) return A;
+  let eoNum = [];
+  for (let i = 0; i < len; i++) {
+    if (A[i] % 2 == 1) {
+      eoNum.push(A[i]);
+    }
+    if (A[i] % 2 == 0) {
+      eoNum.unshift(A[i]);
+    }
+  }
+  return eoNum;
+};
+// 双指针做法,无需新数组，push和unshift操作，利用第三个变量进行交换
+var sortArrayByParity = function (A) {
+  const len = A.length;
+  if (len == 1) return A;
+  let i = 0,
+    j = len - 1;
+  while (i < j) {
+    if ((A[i] % 2 == 1) && (A[j] % 2 == 0)) {
+      let temp = A[j];
+      A[j] = A[i];
+      A[i] = temp;
+      i++;
+      j--;
+    }
+    if (A[i] % 2 == 0) i++;
+    if (A[j] % 2 == 1) j--;
+  }
+  return A;
+};

solution/0985.Sum of Even Numbers After Queries/Solution.js

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+/**
+ * @param {number[]} A
+ * @param {number[][]} queries
+ * @return {number[]}
+ */
+var sumEvenAfterQueries = function (A, queries) {
+  const len = A.length; // A数组的长度
+  const qlen = queries.length; // queries数组的长度
+  let answer = [];
+  let S = 0;
+  for (let i = 0; i < len; i++) {
+    if (A[i] % 2 == 0) {
+      S += A[i];
+    }
+  }
+  for (let j = 0; j < qlen; j++) {
+    let val = queries[j][0];
+    let index = queries[j][1];
+    if (A[index] % 2 == 0) {
+      S -= A[index];
+    }
+    A[index] += val
+    if (A[index] % 2 == 0) {
+      S += A[index]
+
+    }
+    answer.push(S);
+  }
+  return answer;
+};