dp~Leetcode 377. Combination Sum IV

377. Combination Sum IV

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.

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

简析：

dp[i]表示到达i的路径条数。

如数组[1,2,3]

dp[4]=dp[4-1]+dp[4-2]+dp[4-3];

表示到达4的路径条数 = 到达3的路径条数（dp[4-1]）+到达2的路径条数（dp[4-2]）+到达1的路径条数（dp[4-3]）

状态方程：

1、i>nums[j]时

dp[i]=dp[i-nums[0]]+dp[i-nums[1]]+dp[i-nums[2]]+......dp[i-nums[j]]+...

2、i==nums[j]时

dp[i]=dp[i]+1;//如数组[1,2,3]---->当j=2,即nums[2]=3时，dp[3]=dp[3]+1,从3到3只有一条路径，所以加1即可！