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  • 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.
    

    Follow up:
    What if negative numbers are allowed in the given array?
    How does it change the problem?
    What limitation we need to add to the question to allow negative numbers?

    Analyse: For every number i from 1 ... target, we check the numbers num in nums, if i >= num, we seperate i into num and i - num, and compute how many ways i - num are consisted. That is to say, dp[i] += dp[i - num].

    Runtime: 4ms. 

     1 class Solution {
     2 public:
     3     int combinationSum4(vector<int>& nums, int target) {
     4         vector<int> dp(target + 1, 0);
     5         
     6         dp[0] = 1;
     7         for(int i = 1; i <= target; i++) {
     8             for(auto num : nums) {
     9                 if(i >= num)
    10                     dp[i] += dp[i - num];
    11             }
    12         }
    13         return dp[target];
    14     }
    15 };
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  • 原文地址:https://www.cnblogs.com/amazingzoe/p/5745334.html
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