@@ -345,6 +345,76 @@ var threeSum = function(nums) {
345345 return res
346346};
347347```
348+ 349+ 解法二:nSum通用解法。递归
350+ 351+ ``` js
352+ /**
353+ * nsum通用解法,支持2sum,3sum,4sum...等等
354+ * 时间复杂度分析:
355+ * 1. n = 2时,时间复杂度O(NlogN),排序所消耗的时间。、
356+ * 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)
357+ * @param {number[]} nums
358+ * @return {number[][]}
359+ */
360+ var threeSum = function (nums ) {
361+ // nsum通用解法核心方法
362+ function nSumTarget (nums , n , start , target ) {
363+ // 前提:nums要先排序好
364+ let res = [];
365+ if (n === 2 ) {
366+ res = towSumTarget (nums, start, target);
367+ } else {
368+ for (let i = start; i < nums .length ; i++ ) {
369+ // 递归求(n - 1)sum
370+ let subRes = nSumTarget (
371+ nums,
372+ n - 1 ,
373+ i + 1 ,
374+ target - nums[i]
375+ );
376+ for (let j = 0 ; j < subRes .length ; j++ ) {
377+ res .push ([nums[i], ... subRes[j]]);
378+ }
379+ // 跳过相同元素
380+ while (nums[i] === nums[i + 1 ]) i++ ;
381+ }
382+ }
383+ return res;
384+ }
385+ 386+ function towSumTarget (nums , start , target ) {
387+ // 前提:nums要先排序好
388+ let res = [];
389+ let len = nums .length ;
390+ let left = start;
391+ let right = len - 1 ;
392+ while (left < right) {
393+ let sum = nums[left] + nums[right];
394+ if (sum < target) {
395+ while (nums[left] === nums[left + 1 ]) left++ ;
396+ left++ ;
397+ } else if (sum > target) {
398+ while (nums[right] === nums[right - 1 ]) right-- ;
399+ right-- ;
400+ } else {
401+ // 相等
402+ res .push ([nums[left], nums[right]]);
403+ // 跳过相同元素
404+ while (nums[left] === nums[left + 1 ]) left++ ;
405+ while (nums[right] === nums[right - 1 ]) right-- ;
406+ left++ ;
407+ right-- ;
408+ }
409+ }
410+ return res;
411+ }
412+ nums .sort ((a , b ) => a - b);
413+ // n = 3,此时求3sum之和
414+ return nSumTarget (nums, 3 , 0 , 0 );
415+ };
416+ ```
417+ 348418TypeScript:
349419
350420``` typescript
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