LeetCode 494. Target Sum

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本文探讨了一个经典的算法问题：给定一组非负整数和一个目标值S，通过选择每个整数前的符号（+ 或 -），找出使整数之和等于目标值S的所有可能方式的数量。文章提供了一个Python实现的解决方案，利用动态规划思想，遍历所有可能的组合，最终返回符合条件的组合数量。

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.

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注意边界条件

class Solution:
    def findTargetSumWays(self, nums: List[int], S: int) -> int:
        l,c,n = len(nums),0,1
        if (l <= 0):
            return 0
        layers = [{-nums[0]:1,nums[0]:1},{}] if (nums[0] != 0) else [{0:2},{}]
        for i in range(1,l):
            for kk,vv in layers[c].items():
                for dk in [-nums[i],nums[i]]:
                    newk = kk+dk
                    newv = layers[n][newk] if newk in layers[n] else 0
                    layers[n][newk] = newv+vv
            layers[c].clear()
            c,n = n,c
        return layers[c][S] if S in layers[c] else 0