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Merge branch 'youngyangyang04:master' into master
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@ -646,5 +646,68 @@ func countNodes(_ root: TreeNode?) -> Int {
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}
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```
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## Scala
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递归:
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```scala
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object Solution {
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def countNodes(root: TreeNode): Int = {
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if(root == null) return 0
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1 + countNodes(root.left) + countNodes(root.right)
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}
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}
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```
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层序遍历:
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```scala
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object Solution {
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import scala.collection.mutable
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def countNodes(root: TreeNode): Int = {
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if (root == null) return 0
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val queue = mutable.Queue[TreeNode]()
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var node = 0
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queue.enqueue(root)
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while (!queue.isEmpty) {
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val len = queue.size
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for (i <- 0 until len) {
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node += 1
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val curNode = queue.dequeue()
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if (curNode.left != null) queue.enqueue(curNode.left)
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if (curNode.right != null) queue.enqueue(curNode.right)
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}
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}
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node
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}
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}
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```
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利用完全二叉树性质:
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```scala
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object Solution {
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def countNodes(root: TreeNode): Int = {
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if (root == null) return 0
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var leftNode = root.left
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var rightNode = root.right
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// 向左向右往下探
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var leftDepth = 0
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while (leftNode != null) {
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leftDepth += 1
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leftNode = leftNode.left
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}
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var rightDepth = 0
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while (rightNode != null) {
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rightDepth += 1
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rightNode = rightNode.right
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}
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// 如果相等就是一个满二叉树
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if (leftDepth == rightDepth) {
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return (2 << leftDepth) - 1
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}
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// 如果不相等就不是一个完全二叉树,继续向下递归
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countNodes(root.left) + countNodes(root.right) + 1
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}
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}
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```
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-----------------------
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<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>
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@ -702,5 +702,35 @@ func binaryTreePaths(_ root: TreeNode?) -> [String] {
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}
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```
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Scala:
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递归:
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```scala
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object Solution {
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import scala.collection.mutable.ListBuffer
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def binaryTreePaths(root: TreeNode): List[String] = {
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val res = ListBuffer[String]()
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def traversal(curNode: TreeNode, path: ListBuffer[Int]): Unit = {
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path.append(curNode.value)
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if (curNode.left == null && curNode.right == null) {
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res.append(path.mkString("->")) // mkString函数: 将数组的所有值按照指定字符串拼接
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return // 处理完可以直接return
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}
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if (curNode.left != null) {
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traversal(curNode.left, path)
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path.remove(path.size - 1)
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}
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if (curNode.right != null) {
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traversal(curNode.right, path)
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path.remove(path.size - 1)
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}
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}
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traversal(root, ListBuffer[Int]())
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res.toList
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}
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}
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```
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-----------------------
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<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>
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@ -417,6 +417,26 @@ var canPartition = function(nums) {
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```
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TypeScript:
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```ts
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function canPartition(nums: number[]): boolean {
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const sum: number = nums.reduce((a: number, b: number): number => a + b);
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if (sum % 2 === 1) return false;
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const target: number = sum / 2;
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// dp[j]表示容量(总数和)为j的背包所能装下的数(下标[0, i]之间任意取)的总和(<= 容量)的最大值
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const dp: number[] = new Array(target + 1).fill(0);
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const n: number = nums.length;
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for (let i: number = 0; i < n; i++) {
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for (let j: number = target; j >= nums[i]; j--) {
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dp[j] = Math.max(dp[j], dp[j - nums[i]] + nums[i]);
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}
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}
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return dp[target] === target;
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};
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```
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C:
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二维dp:
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```c
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@ -575,6 +595,5 @@ function canPartition(nums: number[]): boolean {
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-----------------------
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<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>
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@ -351,22 +351,26 @@ const findTargetSumWays = (nums, target) => {
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};
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```
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TypeScript:
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```typescript
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TypeScript:
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```ts
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function findTargetSumWays(nums: number[], target: number): number {
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const sum: number = nums.reduce((pre, cur) => pre + cur);
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if (Math.abs(target) > sum) return 0;
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if ((target + sum) % 2 === 1) return 0;
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const bagSize: number = (target + sum) / 2;
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const dp: number[] = new Array(bagSize + 1).fill(0);
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dp[0] = 1;
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for (let i = 0; i < nums.length; i++) {
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for (let j = bagSize; j >= nums[i]; j--) {
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// 把数组分成两个组合left, right.left + right = sum, left - right = target.
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const sum: number = nums.reduce((a: number, b: number): number => a + b);
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if ((sum + target) % 2 || Math.abs(target) > sum) return 0;
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const left: number = (sum + target) / 2;
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// 将问题转化为装满容量为left的背包有多少种方法
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// dp[i]表示装满容量为i的背包有多少种方法
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const dp: number[] = new Array(left + 1).fill(0);
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dp[0] = 1; // 装满容量为0的背包有1种方法(什么也不装)
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for (let i: number = 0; i < nums.length; i++) {
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for (let j: number = left; j >= nums[i]; j--) {
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dp[j] += dp[j - nums[i]];
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}
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}
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return dp[bagSize];
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return dp[left];
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};
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```
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@ -277,26 +277,29 @@ var lastStoneWeightII = function (stones) {
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};
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```
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TypeScript:
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```typescript
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TypeScript版本
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```ts
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function lastStoneWeightII(stones: number[]): number {
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const sum: number = stones.reduce((pre, cur) => pre + cur);
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const bagSize: number = Math.floor(sum / 2);
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const weightArr: number[] = stones;
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const valueArr: number[] = stones;
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const goodsNum: number = weightArr.length;
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const dp: number[] = new Array(bagSize + 1).fill(0);
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for (let i = 0; i < goodsNum; i++) {
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for (let j = bagSize; j >= weightArr[i]; j--) {
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dp[j] = Math.max(dp[j], dp[j - weightArr[i]] + valueArr[i]);
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const sum: number = stones.reduce((a: number, b:number): number => a + b);
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const target: number = Math.floor(sum / 2);
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const n: number = stones.length;
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// dp[j]表示容量(总数和)为j的背包所能装下的数(下标[0, i]之间任意取)的总和(<= 容量)的最大值
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const dp: number[] = new Array(target + 1).fill(0);
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for (let i: number = 0; i < n; i++ ) {
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for (let j: number = target; j >= stones[i]; j--) {
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dp[j] = Math.max(dp[j], dp[j - stones[i]] + stones[i]);
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}
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}
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return sum - dp[bagSize] * 2;
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return sum - dp[target] - dp[target];
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};
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```
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-----------------------
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<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>
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@ -217,6 +217,46 @@ var smallerNumbersThanCurrent = function(nums) {
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};
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```
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TypeScript:
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> 暴力法:
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```typescript
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function smallerNumbersThanCurrent(nums: number[]): number[] {
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const length: number = nums.length;
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const resArr: number[] = [];
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for (let i = 0; i < length; i++) {
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let count: number = 0;
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for (let j = 0; j < length; j++) {
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if (nums[j] < nums[i]) {
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count++;
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}
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}
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resArr[i] = count;
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}
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return resArr;
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};
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```
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> 排序+hash
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```typescript
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function smallerNumbersThanCurrent(nums: number[]): number[] {
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const length: number = nums.length;
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const sortedArr: number[] = [...nums];
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sortedArr.sort((a, b) => a - b);
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const hashMap: Map<number, number> = new Map();
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for (let i = length - 1; i >= 0; i--) {
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hashMap.set(sortedArr[i], i);
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}
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const resArr: number[] = [];
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for (let i = 0; i < length; i++) {
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resArr[i] = hashMap.get(nums[i]);
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}
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return resArr;
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};
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```
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-----------------------
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Reference in New Issue
Block a user