k Sum | & ||

k Sum

Given n distinct positive integers, integer k (k <= n) and a number target.

Find k numbers where sum is target. Calculate how many solutions there are?

Example

Given [1,2,3,4], k = 2, target = 5.

There are 2 solutions: [1,4] and [2,3].

Return 2.

分析：

第一种方法用递归，但是超时了。

 public class Solution {
     public int kSum(int A[], int k, int target) {
         int[] total = new int[];
         helper(A, , k, , target, , total);
         return total[];
     }

     public void helper(int[] A, int index, int k, int count, int target, int total, int[] kk) {
         if (count > k || index >= A.length || total > target) return;

         total += A[index];
         count++;

         if (count == k && total == target) {
             kk[]++;
         }

         helper(A, index + , k, count, target, total, kk);
         total -= A[index];
         count--;
         helper(A, index + , k, count, target, total, kk);
     }
 }

很明显，the preferred approach is DP. 但是如何做呢？我做不出来。 :-( 还是直接copy paste其它牛人的解答吧。

 

 F[0][0][0]表示在一个空集中找出0个数，target为0，则有1个解，就是什么也不挑嘛！ 其实应该这样写，也就是说，找0个数，目标为0，则一定是有1个解：

if (j == 0 && t == 0) {
　　// select 0 number from i to the target: 0
　　D[i][j][t] = 1;
}

1. 状态表达式：

D[i][j][t] = D[i - 1][j][t];
if (t - A[i - 1] >= 0) {
D[i][j][t] += D[i - 1][j - 1][t - A[i - 1]];
}

意思就是：

（1）我们可以把当前A[i - 1]这个值包括进来，所以需要加上D[i - 1][j - 1][t - A[i - 1]]（前提是t - A[i - 1]要大于0）

（2）我们可以不选择A[i - 1]这个值，这种情况就是D[i - 1][j][t]，也就是说直接在前i-1个值里选择一些值加到target.

 public class Solution {
     public int  kSum(int A[], int k, int target) {
         if (target < ) return ;
         int len = A.length;
         int[][][] D = new int[len + ][k + ][target + ];

         for (int i = ; i <= len; i++) {
             for (int j = ; j <= k; j++) {
                 for (int t = ; t <= target; t++) {
                     if (j ==  && t == ) {
                         // select 0 number from i to the target: 0
                         D[i][j][t] = ;
                     } else if (!(i ==  || j ==  || t == )) {
                         D[i][j][t] = D[i - ][j][t];
                         if (t - A[i - ] >= ) {
                             D[i][j][t] += D[i - ][j - ][t - A[i - ]];
                         }
                     }
                 }
             }
         }
         return D[len][k][target];
     }
 }

k Sum II

Given n unique integers, number k (1<=k<=n) and target.

Find all possible k integers where their sum is target.

Have you met this question in a real interview?

Yes

Example

Given [1,2,3,4], k = 2, target = 5. Return:

[
  [1,4],
  [2,3]
]

 public class Solution {

     public ArrayList<ArrayList<Integer>> kSumII(int[] A, int k, int target) {
         ArrayList<ArrayList<Integer>> allList = new ArrayList<ArrayList<Integer>>();
         ArrayList<Integer> list = new ArrayList<Integer>();
         if (A == null || A.length ==  || k == ) return allList;

         helper(allList, list, , A, k, , target, );
         return allList;
     }

     public void helper(ArrayList<ArrayList<Integer>> allList, ArrayList<Integer> list, int index, int[] A, int k, int count, int target, int total) {
         if (count > k || index >= A.length || total > target) return;

         list.add(A[index]);
         total += A[index];
         count++;

         if (count == k && total == target) {
             allList.add(new ArrayList<Integer>(list));
         }

         helper(allList, list, index + , A, k, count, target, total);
         total -= list.get(list.size() - );
         list.remove(list.size() - );
         count--;
         helper(allList, list, index + , A, k, count, target, total);
     }
 }

Reference:

http://www.cnblogs.com/yuzhangcmu/p/4279676.html