[Leetcode] #416 Partition Equal Subset Sum

Discription:

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.

Example 1:

Input: [1, 5, 11, 5]

Output: true

Explanation: The array can be partitioned as [1, 5, 5] and [11].

Example 2:

Input: [1, 2, 3, 5]

Output: false

Explanation: The array cannot be partitioned into equal sum subsets.

Solution:

01背包问题。

bool canPartition(vector<int>& nums) {
int sum = 0;
int n = nums.size();
for (int i = 0; i < nums.size(); i++){
sum += nums[i];
}
if (sum % 2 == 1)
return false;
sum = sum / 2;
vector<vector<int>> dp(n + 1, vector<int>(sum + 1, 0));
for (int i = 1; i <= n; i++){
for (int j = 1; j <= sum; j++){
if (j >= nums[i - 1]){
dp[i][j] = max(dp[i - 1][j], dp[i - 1][j - nums[i - 1]] + nums[i - 1]);
}
else{
dp[i][j] = dp[i - 1][j];
}
}
}
return dp
[sum] == sum ? true : false;
}

bool canPartition(vector<int>& nums) { //利用滚动数组实现
int sum = 0;
int n = nums.size();
for (int i = 0; i < nums.size(); i++){
sum += nums[i];
}
if (sum % 2 == 1)
return false;
sum = sum / 2;
vector<int> dp(sum + 1, 0);
for (int i = 1; i <= n; i++){
for (int j = sum; j >= nums[i-1]; j--)
dp[j] = max(dp[j], dp[j - nums[i - 1]] + nums[i - 1]);
}
return dp[sum] == sum ? true : false;
}

GitHub-Leetcode:https://github.com/wenwu313/LeetCode