lc 494 Target Sum
You are given a list of non-negative integers, a1, a2, ..., an, and a target, S. Now you have 2 symbols + and -. For each integer, you should choose one from + and - as its new symbol.
Find out how many ways to assign symbols to make sum of integers equal to target S.
Example 1:
Input: nums is [1, 1, 1, 1, 1], S is 3.
Output: 5
Explanation:
-1+1+1+1+1 = 3
+1-1+1+1+1 = 3
+1+1-1+1+1 = 3
+1+1+1-1+1 = 3
+1+1+1+1-1 = 3
There are 5 ways to assign symbols to make the sum of nums be target 3.
Note:
- The length of the given array is positive and will not exceed 20.
- The sum of elements in the given array will not exceed 1000.
- Your output answer is guaranteed to be fitted in a 32-bit integer.
DP Accepted
数组的数字要么取正要么取负,sum(P)代表所有取正的数的和,sum(N)代表所有取负的数的和。
sum(P) - sum(N) = target
sum(P) + sum(N) + sum(P) - sum(N) = target + sum(P) + sum(N)
2 * sum(P) = target + sum(nums)
可以看出:如果target + sum(nums)为奇数,那必定可行方法数为0。之后利用 416 Partition Equal Subset Sum中的方法,即可求得数组中求和得到(target + sum(nums))/2的方法数。
class Solution {
public:
int findTargetSumWays(vector<int>& nums, int S) {
int sum = accumulate(nums.begin(), nums.end(), 0);
if (sum < S) return 0;
else return (S+sum) & 1 ? 0 : subsum(nums, (S+sum)>>1);
}
int subsum(vector<int>& nums, int S) {
vector<int> dp(S+1, 0);
dp[0] = 1;
for (int n : nums)
for (int i = S; i >= n; i--)
dp[i] += dp[i-n];
return dp[S];
}
};