(算法分析Week13)Combination Sum IV[Medium]

本文链接：https://blog.youkuaiyun.com/qq_32335095/article/details/78692187

本文探讨了组合总和IV问题的解决方案，采用动态规划方法找出所有可能的组合方式，以满足给定的目标值。通过具体示例展示了如何计算不同序列的组合数量，并提供了详细的代码实现。

506. 377. Combination Sum IV[Medium]

Description

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.

Solution

动态规划，有点类似跳楼梯那道题，只不过跳楼梯规定了0或1，这里有vector中的数字供选取，其实思路还是一样的。
dp[i]表示目标数为i的解的个数，从target-nums[i]到target只有一种途径，所以这种途径的可能是dp[target-nums[i]]，
也就是：

dp[i]=Σdp[i-nums[k]]  0<=k<=nums.size()

当i-nums[k]等于0时，表示数组中有target，此时dp[i]为1

Complexity analysis

O(MN)
M = target;
N = nums.size()

Code

class Solution {
public:
    int combinationSum4(vector<int>& nums, int target) {
        int dp[target+1] = {0};
            for (int i = 1; i <= target; i++) {
            for (int j = 0; j < nums.size(); j++) {
                if (i - nums[j] > 0) {
                    dp[i] += dp[i - nums[j]];
                } else if (i - nums[j] == 0) {
                    dp[i] += 1;
                }
            }
        }
        return dp[target];
    }
};

Result

这里写图片描述