zoukankan      html  css  js  c++  java
  • LN : leetcode 494 Target Sum

    lc 494 Target Sum


    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.

    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];
        }
    };
    
  • 相关阅读:
    yum和apt-get用法及区别
    Vim终极指南:所思即所得
    Vim 操作符命令和动作命令
    vim中文手册
    标准代码页(codepage)列表
    中文字符集编码Unicode ,gb2312 , cp936 ,GBK,GB18030
    vim在系统剪切板的复制与粘贴
    linux中高亮显示文本的工具 -- bat
    pandas category数据类型
    pandas 数据类型转换
  • 原文地址:https://www.cnblogs.com/renleimlj/p/8025194.html
Copyright © 2011-2022 走看看