Merge branch 'youngyangyang04:master' into master

This commit is contained in:
yqq
2021-09-19 18:17:39 +08:00
committed by GitHub
4 changed files with 41 additions and 20 deletions

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@ -185,7 +185,7 @@ func removeElement(nums []int, val int) int {
```
JavaScript:
```
```javascript
//时间复杂度O(n)
//空间复杂度O(1)
var removeElement = (nums, val) => {

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@ -159,26 +159,28 @@ func getSum(n int) int {
javaScript:
```js
function getN(n) {
if (n == 1 || n == 0) return n;
let res = 0;
while (n) {
res += (n % 10) * (n % 10);
n = parseInt(n / 10);
var isHappy = function (n) {
let m = new Map()
const getSum = (num) => {
let sum = 0
while (n) {
sum += (n % 10) ** 2
n = Math.floor(n / 10)
}
return sum
}
while (true) {
// n出现过证明已陷入无限循环
if (m.has(n)) return false
if (n === 1) return true
m.set(n, 1)
n = getSum(n)
}
return res;
}
var isHappy = function(n) {
const sumSet = new Set();
while (n != 1 && !sumSet.has(n)) {
sumSet.add(n);
n = getN(n);
}
return n == 1;
};
// 使用环形链表的思想 说明出现闭环 退出循环
// 方法二:使用环形链表的思想 说明出现闭环 退出循环
var isHappy = function(n) {
if (getN(n) == 1) return true;
let a = getN(n), b = getN(getN(n));

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@ -278,7 +278,26 @@ const findLength = (A, B) => {
return res;
};
```
> 滚动数组
```javascript
const findLength = (nums1, nums2) => {
let len1 = nums1.length, len2 = nums2.length;
// dp[i][j]: 以nums1[i-1]、nums2[j-1]为结尾的最长公共子数组的长度
let dp = new Array(len2+1).fill(0);
let res = 0;
for (let i = 1; i <= len1; i++) {
for (let j = len2; j > 0; j--) {
if (nums1[i-1] === nums2[j-1]) {
dp[j] = dp[j-1] + 1;
} else {
dp[j] = 0;
}
res = Math.max(res, dp[j]);
}
}
return res;
}
```
-----------------------

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@ -88,7 +88,7 @@ int main() {
**所以可以看出在C++中二维数组在地址空间上是连续的**
像Java是没有指针的同时也不对程序员暴其元素的地址,寻址操作完全交给虚拟机。
像Java是没有指针的同时也不对程序员暴其元素的地址,寻址操作完全交给虚拟机。
所以看不到每个元素的地址情况这里我以Java为例也做一个实验。