Partition Equal Subset Sum问题及解法
2017-10-18 12:21
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问题描述:
Given a non-empty array containing only positive integers, find if the array can be partitioned into two subsets such that the sum of elements in both subsets is equal.
Note:
Each of the array element will not exceed 100.
The array size will not exceed 200.
示例:
问题分析:
此类问题类似于在数组中查找子序列和为target的题目,很典型的一种动态规划问题。
过程详见代码:
class Solution {
public:
bool canPartition(vector<int>& nums) {
if (nums.size() < 2) return false;
int sum = accumulate(nums.begin(), nums.end(), 0);
if (sum % 2) return false;
vector<bool> dp(sum / 2 + 1, 0);
dp[0] = true;
int target = sum / 2;
for (auto num : nums)
{
for (int i = target; i > 0; i--)
{
if (i >= num)
{
dp[i] = dp[i] || dp[i - num];
}
}
}
return dp[target];
}
};
Given a non-empty array containing only positive integers, find if the array can be partitioned into two subsets such that the sum of elements in both subsets is equal.
Note:
Each of the array element will not exceed 100.
The array size will not exceed 200.
示例:
Input: [1, 5, 11, 5] Output: true Explanation: The array can be partitioned as [1, 5, 5] and [11].
Input: [1, 2, 3, 5] Output: false Explanation: The array cannot be partitioned into equal sum subsets.
问题分析:
此类问题类似于在数组中查找子序列和为target的题目,很典型的一种动态规划问题。
过程详见代码:
class Solution {
public:
bool canPartition(vector<int>& nums) {
if (nums.size() < 2) return false;
int sum = accumulate(nums.begin(), nums.end(), 0);
if (sum % 2) return false;
vector<bool> dp(sum / 2 + 1, 0);
dp[0] = true;
int target = sum / 2;
for (auto num : nums)
{
for (int i = target; i > 0; i--)
{
if (i >= num)
{
dp[i] = dp[i] || dp[i - num];
}
}
}
return dp[target];
}
};
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