[LintCode] Backpack VI 背包之六

Given an integer array nums with all positive numbers and no duplicates, find the number of possible combinations that add up to a positive integer target.

Notice

The different sequences are counted as different combinations.

Have you met this question in a real interview?

Yes

Example

Given nums = [1, 2, 4], target = 4

The possible combination ways are:
[1, 1, 1, 1]
[1, 1, 2]
[1, 2, 1]
[2, 1, 1]
[2, 2]
[4]

return 6

不太懂这题名称为啥叫backpack，LeetCode上的原题，请参见我之前的博客Combination Sum IV 。

解法一：

class Solution {
public:
    /**
     * @param nums an integer array and all positive numbers, no duplicates
     * @param target an integer
     * @return an integer
     */
    int backPackVI(vector<int>& nums, int target) {
        vector<int> dp(target + , );
        dp[] = ;
        for (int i = ; i <= target; ++i) {
            for (auto a : nums) {
                if (a <= i) {
                    dp[i] += dp[i - a];
                }
            }
        }
        return dp.back();
    }
};

解法二：

class Solution {
public:
    /**
     * @param nums an integer array and all positive numbers, no duplicates
     * @param target an integer
     * @return an integer
     */
    int backPackVI(vector<int>& nums, int target) {
        vector<int> dp(target + , );
        dp[] = ;
        sort(nums.begin(), nums.end());
        for (int i = ; i <= target; ++i) {
            for (auto a : nums) {
                if (a > i) break;
                dp[i] += dp[i - a];
            }
        }
        return dp.back();
    }
};