Merge pull request #1403 from Damon0820/master

添加(0015.三数之和.md):增加javascript版本nsum的通用解法
This commit is contained in:
程序员Carl
2022-06-28 11:19:37 +08:00
committed by GitHub
2 changed files with 88 additions and 0 deletions

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@ -345,6 +345,76 @@ var threeSum = function(nums) {
return res
};
```
解法二nSum通用解法递归
```js
/**
* nsum通用解法支持2sum3sum4sum...等等
* 时间复杂度分析:
* 1. n = 2时时间复杂度O(NlogN),排序所消耗的时间。、
* 2. n > 2时时间复杂度为O(N^n-1)即N的n-1次方至少是2次方此时可省略排序所消耗的时间。举例3sum为O(n^2)4sum为O(n^3)
* @param {number[]} nums
* @return {number[][]}
*/
var threeSum = function (nums) {
// nsum通用解法核心方法
function nSumTarget(nums, n, start, target) {
// 前提nums要先排序好
let res = [];
if (n === 2) {
res = towSumTarget(nums, start, target);
} else {
for (let i = start; i < nums.length; i++) {
// 递归求(n - 1)sum
let subRes = nSumTarget(
nums,
n - 1,
i + 1,
target - nums[i]
);
for (let j = 0; j < subRes.length; j++) {
res.push([nums[i], ...subRes[j]]);
}
// 跳过相同元素
while (nums[i] === nums[i + 1]) i++;
}
}
return res;
}
function towSumTarget(nums, start, target) {
// 前提nums要先排序好
let res = [];
let len = nums.length;
let left = start;
let right = len - 1;
while (left < right) {
let sum = nums[left] + nums[right];
if (sum < target) {
while (nums[left] === nums[left + 1]) left++;
left++;
} else if (sum > target) {
while (nums[right] === nums[right - 1]) right--;
right--;
} else {
// 相等
res.push([nums[left], nums[right]]);
// 跳过相同元素
while (nums[left] === nums[left + 1]) left++;
while (nums[right] === nums[right - 1]) right--;
left++;
right--;
}
}
return res;
}
nums.sort((a, b) => a - b);
// n = 3此时求3sum之和
return nSumTarget(nums, 3, 0, 0);
};
```
TypeScript:
```typescript

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@ -234,6 +234,7 @@ func fib(n int) int {
}
```
### Javascript
解法一
```Javascript
var fib = function(n) {
let dp = [0, 1]
@ -244,6 +245,23 @@ var fib = function(n) {
return dp[n]
};
```
解法二时间复杂度O(N)空间复杂度O(1)
```Javascript
var fib = function(n) {
// 动规状态转移中当前结果只依赖前两个元素的结果所以只要两个变量代替dp数组记录状态过程。将空间复杂度降到O(1)
let pre1 = 1
let pre2 = 0
let temp
if (n === 0) return 0
if (n === 1) return 1
for(let i = 2; i <= n; i++) {
temp = pre1
pre1 = pre1 + pre2
pre2 = temp
}
return pre1
};
```
TypeScript