[LeetCode]Target sum

题目描述:

LeetCode 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:

  1. The length of the given array is positive and will not exceed 20.
  2. The sum of elements in the given array will not exceed 1000.
  3. Your output answer is guaranteed to be fitted in a 32-bit integer.

题目大意:

给定一组非负整数a1, a2, ..., an,以及一个目标数S。给定两种符号+和-,对于每一个整数,选择一个运算符。

计算有多少种运算符的组合方式,可以使整数的和为目标数S。

注意:

  • 给定数组长度为正数并且不会超过20。
  • 元素之和不会超过1000。
  • 输出答案确保在32位整数范围内。

解题思路:

动态规划(Dynamic Programming)

状态转移方程:dp[i + 1][k + nums[i] * sgn] += dp[i][k]

上式中,sgn取值±1,k为dp[i]中保存的所有状态;初始令dp[0][0] = 1

利用滚动数组,可以将空间复杂度优化到O(n),n为可能的运算结果的个数

Python代码:

class Solution(object):
    def findTargetSumWays(self, nums, S):
        """
        :type nums: List[int]
        :type S: int
        :rtype: int
        """
        dp = collections.Counter()
        dp[0] = 1
        for n in nums:
            ndp = collections.Counter()
            for sgn in (1, -1):
                for k in dp.keys():
                    ndp[k + n * sgn] += dp[k]
            dp = ndp
        return dp[S]

 

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