leetcode377.Combination Sum IV

发表于 更新于 分类于 ，

题目要求

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
Given an integer array with all positive numbers and no duplicates, find the number of possible combinations that add up to a positive integer target.

Example:

nums = [1, 2, 3]
target = 4

The possible combination ways are:
(1, 1, 1, 1)
(1, 1, 2)
(1, 2, 1)
(1, 3)
(2, 1, 1)
(2, 2)
(3, 1)

Note that different sequences are counted as different combinations.

Therefore the output is 7.

Follow up:
What if negative numbers are allowed in the given array?
How does it change the problem?
What limitation we need to add to the question to allow negative numbers?

有一个不包含重复值的正整数数组nums，问从数组中选择几个数，其和为target，这样的数的组合有几种？

思路一：自顶向下的dp

这题本质上需要注意一点，就是我如果需要组成target，那么一定是由nums中的一个值和另一个值的排列组合结果构成的。比如com[4] = com[4-1] + com[4-2] + com[4-1]。通过这一点，我们构成一个递归表达式，但是因为单纯的递归表达式没有计算中间结果，所以会造成大量重复的计算影响效率，所以这里采用dp的思路额外的用数组来记录已经计算过的com结果。比如com[3] = com[2] + com[1], com[2] = com[1]，如果没有dp，则需要重复计算com[1]的结果。

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
public int combinationSum4(int[] nums, int target) {
     if (nums == null || nums.length < 1) {
         return 0;
     }
     int[] dp = new int[target + 1];
     Arrays.fill(dp, -1);
     dp[0] = 1;
     return helper(nums, target, dp);
 }
 
 private int helper(int[] nums, int target, int[] dp) {
     if (dp[target] != -1) {
         return dp[target];
     }
     int res = 0;
     for (int i = 0; i < nums.length; i++) {
         if (target >= nums[i]) {
             res += helper(nums, target - nums[i], dp);
         }
     }
     dp[target] = res;
     return res;
 }

思路二：自底向上dp

1
2
3
4
5
6
7
8
9
10
11
12
13
public int combinationSum4(int[] nums, int target) {
    Arrays.sort(nums);
    int[] combinationCount = new int[target+1];
    
    combinationCount[0] = 1;
    
    for(int i = 1 ; i<=target ; i++) {
        for(int j = 0 ; j<nums.length && nums[j] <= i ; j++) {
            combinationCount[i] += combinationCount[i - nums[j]];
        }
    }
    return combinationCount[target];
}