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?题目含义:给定多个正整数且没有重复,从中挑选任意个数字(可以重复)使其加起来等于给定目标值target
思路:
dp[i]表示总和为i的情况数,假设当前整数值等于n(n<i),如果我们知道了 dp[i-n] 的情况数,因为 dp[i-n] 的每一种情况加n都等于dp[i],所以说dp[i-n]和dp[i]的情况数相同。
每获取一个整数n,都会给了dp[i]添加dp[i-n]种情况,所以公式为dp[i] = dp[i] + dp[i-n]例如:
当前元素n等于1时,dp[9] = dp[9] + dp[9-1] dp[8] = dp[8] + dp[8-1] ... dp[1] = dp[1] + dp[1-1] 当前元素n等于2时,dp[9] = dp[9] + dp[9-2] dp[8] = dp[8] + dp[8-2] ... dp[2] = dp[2] + dp[2-2] 当前元素n等于3时,dp[9] = dp[9] + dp[9-3] dp[8] = dp[8] + dp[8-3] ... dp[3] = dp[3] + dp[3-3]1 public int combinationSum4(int[] nums, int target) { 2 int[] dp=new int[target+1]; 3 dp[0]=1; 4 for(int i=1;i<=target;i++) 5 { 6 for (int num:nums) 7 { 8 if(i>=num) dp[i]=dp[i]+dp[i-num]; 9 }10 }11 return dp[target];12 }
类似题目:
请注意:本题目选取的数字可以多次重复,而题目494中的数字不可重复选择