【Leetcode】Combination Sum IV
阿新 • • 發佈:2019-02-04
題目連結:
題目:
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?
思路:
dp,dp[i]表示當target為i 時,有多少種組合。
狀態轉移方程:dp[i]=Σdp[i-nums[k]] 0<=k<=nums.length
當然,需要考慮當i-nums[k]為0時,表示陣列中有target,則此時dp[i]為1,
時間複雜度O(n^2)
演算法:
public int combinationSum4(int[] nums, int target) { int dp[] = new int[target + 1]; for (int i = 0; i <= target; i++) { for (int k = 0; k < nums.length; k++) { if (i - nums[k] > 0) { dp[i] += dp[i - nums[k]]; } else if (i - nums[k] == 0) { dp[i] += 1; } } } return dp[target]; }