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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
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@ -345,6 +345,76 @@ var threeSum = function(nums) {
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return res
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};
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```
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解法二:nSum通用解法。递归
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```js
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/**
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* nsum通用解法,支持2sum,3sum,4sum...等等
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* 时间复杂度分析:
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* 1. n = 2时,时间复杂度O(NlogN),排序所消耗的时间。、
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* 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)
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* @param {number[]} nums
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* @return {number[][]}
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*/
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var threeSum = function (nums) {
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// nsum通用解法核心方法
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function nSumTarget(nums, n, start, target) {
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// 前提:nums要先排序好
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let res = [];
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if (n === 2) {
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res = towSumTarget(nums, start, target);
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} else {
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for (let i = start; i < nums.length; i++) {
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// 递归求(n - 1)sum
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let subRes = nSumTarget(
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nums,
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n - 1,
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i + 1,
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target - nums[i]
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);
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for (let j = 0; j < subRes.length; j++) {
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res.push([nums[i], ...subRes[j]]);
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}
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// 跳过相同元素
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while (nums[i] === nums[i + 1]) i++;
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}
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}
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return res;
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}
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function towSumTarget(nums, start, target) {
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// 前提:nums要先排序好
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let res = [];
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let len = nums.length;
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let left = start;
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let right = len - 1;
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while (left < right) {
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let sum = nums[left] + nums[right];
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if (sum < target) {
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while (nums[left] === nums[left + 1]) left++;
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left++;
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} else if (sum > target) {
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while (nums[right] === nums[right - 1]) right--;
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right--;
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} else {
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// 相等
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res.push([nums[left], nums[right]]);
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// 跳过相同元素
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while (nums[left] === nums[left + 1]) left++;
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while (nums[right] === nums[right - 1]) right--;
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left++;
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right--;
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}
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}
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return res;
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}
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nums.sort((a, b) => a - b);
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// n = 3,此时求3sum之和
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return nSumTarget(nums, 3, 0, 0);
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};
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```
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TypeScript:
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```typescript
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@ -234,6 +234,7 @@ func fib(n int) int {
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}
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```
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### Javascript
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解法一
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```Javascript
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var fib = function(n) {
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let dp = [0, 1]
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@ -244,6 +245,23 @@ var fib = function(n) {
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return dp[n]
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};
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```
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解法二:时间复杂度O(N),空间复杂度O(1)
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```Javascript
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var fib = function(n) {
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// 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)
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let pre1 = 1
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let pre2 = 0
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let temp
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if (n === 0) return 0
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if (n === 1) return 1
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for(let i = 2; i <= n; i++) {
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temp = pre1
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pre1 = pre1 + pre2
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pre2 = temp
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}
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return pre1
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};
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```
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TypeScript
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