Combination Sum IV
早上起來看到朋友微信說出新題了，于是打開leetcode 看了一下，原來又是Combination Sum。 打開之后果斷用了之前三個Combination Sum 用到的 Backtracking, then submit, TIME LIMIT EXCEEDED, XD. 看了下tags，原來是用DP，于是寫了個DP的sulotion

tags

狀態(tài)轉(zhuǎn)移方程: combinationSum4(x): f(x)
f(target) = f(target - nums[0]) + f(target - nums[1]) + ... + f(target - nums.back())

附上之前三個combinationSum的題，沒有做前三個的可以先做一下前三個
Combination Sum
Combination Sum II
Combination Sum III

    int combinationSum4(vector<int>& nums, int target) {
        vector<int> res(target + 1, 0);
        res[0] = 1;
        for (int i = 1; i <= target; i++)
            for (int n: nums)
                if (i >= n)
                    res[i] += res[i - n];
        return res.back();
    }