416. Partition Equal Subset Sum
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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.
是否能把数组分成和相等的两部分
其实也就是求是否能从数组中取出一些数字 使他们的和为sum/2 其中sum是数组全部数字的和
一开始想法有点乱 想到了2sum 对于这道题 需要一直求到(n-1)sum 其中n是数字的个数
后来想到0/1背包问题 对于每一个数字 只有两种选择 取或者不取
public boolean canPartition(int[] nums) { int sum = 0; for (int n : nums) { sum += n; } if (sum % 2 != 0) return false; Arrays.sort(nums); return part(nums, 0, sum/2); } public boolean part(int[] nums, int index, int target) { if (target == 0) return true; if (target < 0) return false; for (int i = index; i < nums.length; i++) { if (part(nums, i+1, target-nums[i]) || part(nums, i+1, target)) return true; } return false; }然而 超时了… 时间复杂度O(n^n)
想到了dynamic programming 但是想不出来最优子结构
以下是discuss中的top1
public boolean canPartition(int[] nums) { int sum = 0; for (int num : nums) { sum += num; } if ((sum & 1) == 1) { return false; } sum /= 2; int n = nums.length; boolean[][] dp = new boolean[n+1][sum+1]; for (int i = 0; i < dp.length; i++) { Arrays.fill(dp[i], false); } dp[0][0] = true; for (int i = 1; i < n+1; i++) { dp[i][0] = true;//任意前i个数字 都可以使sum=0 就是都不取呗 } for (int j = 1; j < sum+1; j++) { dp[0][j] = false;//前0个数字 做不到sum>0 } //对于前i个数字 尝试是否能达成sum=j for (int i = 1; i < n+1; i++) { for (int j = 1; j < sum+1; j++) { dp[i][j] = dp[i-1][j]; if (j >= nums[i-1]) { dp[i][j] = (dp[i][j] || dp[i-1][j-nums[i-1]]); } } } return dp[n][sum];}
其中dp[i][j]表示使用前i个数字 是否能使sum=j
比如输入是1,5,11,5 双层for循环 流程是这样的
先尝试前1个数字 是否能达成sum=1,2,3…11
然后尝试前2个数字 是否能达成sum=1,2,3...11
…
前4个数字 是否能达成sum=1,2,3…11
另外 最优子结构中 dp[i][j] = (dp[i][j] || dp[i-1][j-nums[i-1]]);
i只用到了前一行 也就是i-1的计算结果 所以空间复杂度可以再优化
public boolean canPartition(int[] nums) { int sum = 0; for (int num : nums) { sum += num; } if ((sum & 1) == 1) { return false; } sum /= 2; int n = nums.length; boolean[] dp = new boolean[sum+1]; Arrays.fill(dp, false); dp[0] = true; for (int num : nums) { for (int i = sum; i > 0; i--) { if (i >= num) { dp[i] = dp[i] || dp[i-num]; } } } return dp[sum];}
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- 416. Partition Equal Subset Sum
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- 416. Partition Equal Subset Sum
- 416. Partition Equal Subset Sum
- 416. Partition Equal Subset Sum
- 416. Partition Equal Subset Sum
- 416. Partition Equal Subset Sum
- 416. Partition Equal Subset Sum
- 416. Partition Equal Subset Sum
- 416. Partition Equal Subset Sum
- 416. Partition Equal Subset Sum
- 416. Partition Equal Subset Sum
- 416. Partition Equal Subset Sum
- 416. Partition Equal Subset Sum
- 416. Partition Equal Subset Sum
- 416. Partition Equal Subset Sum
- 416. Partition Equal Subset Sum
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