(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 = 4The 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