15.3sum(Nsum总结)

如果不用hashmap解决2sum，而是把2sum看成nSum中的一部分，那么就需要用two pointers来解决问题。

2sum伪代码：

1. 先对给定的数组进行排序

2. walker指第一个元素，runner指最后一个元素

3.当walker小于runner：

　　1）如果walker和runner的和等于target，那么：

　　　　（1）如果只需要一对解，记录答案，break

　　　　（2）如果需要所有解，那么walker++, runner--

　　2） else如果和小于target，walker++

　　3） else runner++

4.返回结果

算法时间复杂度

1. 2sum的时间复杂度

排序的复杂度是O(NlogN), 然后再就是最多从头到尾走一遍，所以是O(N), 所以复杂度是O(N+NlogN)，所以是O(N)

2. 3sum时间复杂度是

排序的时间复杂度O(NLogN), 在对于每一个数组做2sum，但是这个时候的2sum不需要排序，所以这个2sum是O(N)，所以这部分的时间复杂度是O(N^2）

所以总时间复杂度是O(N^2+NLogN)= O(N^2)

参考链接： http://blog.csdn.net/doc_sgl/article/details/12462151#comments 感谢原博主！

时间： 2024-12-16 16:57:44

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