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

506. 377. Combination Sum IV[Medium]

题目来源

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