377. Combination Sum IV
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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.
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?
Credits:
Special thanks to @pbrother for adding this problem and creating all test cases.
/*
我们需要一个一维数组dp,其中dp[i]表示目标数为i的解的个数,然后我们从1遍历到target,对于每一个数i,遍历nums数组如果i>=x,dp[i] += dp[i - x]。这个也很好理解,比如说对于[1,2,3] 4,这个例子,当我们在计算dp[3]的时候,3可以拆分为1+x,而x即为dp[2],3也可以拆分为2+x,此时x为dp[1],3同样可以拆为3+x,此时x为dp[0],我们把所有的情况加起来就是组成3的所有情况了
*/
public class Solution { public int combinationSum4(int[] nums, int target) { int[] dp = new int[target + 1]; dp[0] = 1; for(int i = 1; i <= target; i++){ for(int j : nums){ if(j <= i){ dp[i] += dp[i - j]; } } } return dp[target]; }}
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- 377. Combination Sum IV
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- 377. Combination Sum IV
- 377. Combination Sum IV
- 377. Combination Sum IV
- 377. Combination Sum IV
- 377. Combination Sum IV
- 377. Combination Sum IV
- 377. Combination Sum IV
- 377. Combination Sum IV
- 377. Combination Sum IV
- 377. Combination Sum IV**
- 377. Combination Sum IV
- 377. Combination Sum IV
- 377. Combination Sum IV
- 377. Combination Sum IV
- 377. Combination Sum IV
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