Nine-chapter algorithm surface question 4 sum

Nine Chapters count judges Net-original website

http://www.jiuzhang.com/problem/69/

TopicsGive a bunch of arrays and a target value, and in this heap of arrays find four numbers that make their sum equal to the target value.

Online test

http://www.lintcode.com/en/problem/4-sum/

Answer

Method One:

The method of this problem is also very similar to the second solution of 2 sum. First we put the array in order first, then we still need four pointers i,j,x,y, we assume i<j<x<y, because we ordered, then A[i]<a[j]<a[x]<a[y], actually test instructions can be converted to a[ I]+a[j]+a[x]+a[y]=v. If we enumerate i,j, then test instructions can be converted to an element greater than a[j] to find a[x]+a[y] = V-a[i]-a[j], then this problem is 2sum, so we can use 2sum Two pointers point to one end of the way to solve the problem.

Method Two:

There is another way to do this problem is to use a hash table to record any of the array of two and sum = A[x]+a[y], and the record is the sum of two numbers, so that the time complexity of O (n^2) can record any two number of sum, and then in two pointers i,j (I<J) Iterate through the array and look for v-a[i]-a[j] in the hash table, and ensure that a[i],a[j],a[x],a[y] cannot take the same number, and then you can find the 4sum by i,j two-layer for loop

Nine-chapter algorithm surface question 4 sum