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279.perfect-squares
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279.perfect-squares.js

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279.perfect-squares.md

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## 题目地址
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https://leetcode.com/problems/perfect-squares/description/
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## 题目描述
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```
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Given a positive integer n, find the least number of perfect square numbers (for example, 1, 4, 9, 16, ...) which sum to n.
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Example 1:
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Input: n = 12
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Output: 3
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Explanation: 12 = 4 + 4 + 4.
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Example 2:
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Input: n = 13
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Output: 2
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Explanation: 13 = 4 + 9.
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```
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## 思路
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直接递归处理即可,但是这种暴力的解法很容易超时。如果你把递归的过程化成一棵树的话(其实就是递归树),
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可以看出中间有很多重复的计算。
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如果能将重复的计算缓存下来,说不定能够解决时间复杂度太高的问题。
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> 递归对内存的要求也很高, 如果数字非常大,也会面临爆栈的风险,将递归转化为循环可以解决。
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如果使用DP,其实本质上和递归 + 缓存 差不多。
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## 关键点解析
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- 如果用递归 + 缓存, 缓存的设计很重要
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我的做法是key就是n,value是以n为起点,到达底端的深度。
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下次取出缓存的时候用当前的level + 存的深度 就是我们想要的level.
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## 代码
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```js
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/*
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* @lc app=leetcode id=279 lang=javascript
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*
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* [279] Perfect Squares
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*
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* https://leetcode.com/problems/perfect-squares/description/
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*
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* algorithms
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* Medium (40.98%)
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* Total Accepted: 168.2K
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* Total Submissions: 408.5K
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* Testcase Example: '12'
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*
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* Given a positive integer n, find the least number of perfect square numbers
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* (for example, 1, 4, 9, 16, ...) which sum to n.
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*
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* Example 1:
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*
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*
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* Input: n = 12
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* Output: 3
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* Explanation: 12 = 4 + 4 + 4.
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*
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* Example 2:
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*
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*
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* Input: n = 13
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* Output: 2
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* Explanation: 13 = 4 + 9.
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*/
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const mapper = {};
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function d(n, level) {
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if (n === 0) return level;
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let i = 1;
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const arr = [];
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while(n - i * i >= 0) {
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const hit = mapper[n - i * i];
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if (hit) {
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arr.push(hit + level);
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} else {
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const depth = d(n - i * i, level + 1) - level;
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mapper[n - i * i] = depth;
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arr.push(depth + level);
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}
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i++;
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}
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return Math.min(...arr);
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}
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/**
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* @param {number} n
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* @return {number}
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*/
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var numSquares = function(n) {
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return d(n, 0);
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};
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```

README.md

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- [328.odd-even-linked-list](./328.odd-even-linked-list.md)
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- [445.add-two-numbers-ii](./445.add-two-numbers-ii.md)
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- [877.stone-game](./877.stone-game.md)
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- [279.perfect-squares](./279.perfect-squares.md)
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### 困难难度
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- [145.binary-tree-postorder-traversal](./145.binary-tree-postorder-traversal.md)

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