2018-08-13 17:35:09
一、Partition Equal Subset Sum
问题描述:
问题求解:
二分和本质上其实是一个背包问题,就是问是否存在一种情况,使得可以填满一个sum/2的背包。
public boolean canPartition(int[] nums) { int sum = 0; for (int i : nums) sum += i; if (sum % 2 != 0) return false; sum /= 2; boolean dp[] = new boolean[sum + 1]; dp[0] = true; for (int num : nums) { for (int i = sum; i >= num; i--) { dp[i] = dp[i] || dp[i - num]; } } return dp[sum]; }
二、Partition to K Equal Sum Subsets
问题描述:
问题求解:
Brute Force,本质是使用组合数进行解空间的遍历。
PS.使用排列数也可以,但是效率比组合数要慢上不少。
public boolean canPartitionKSubsets(int[] nums, int k) { int sum = 0; for (int i = 0; i < nums.length; i++) sum += nums[i]; if (sum % k != 0) return false; Arrays.sort(nums); int eachSum = sum / k; int[] visited = new int[nums.length]; return helper(nums, eachSum, 0, visited, 0, k); } private boolean helper(int[] nums, int eachSum, int curSum, int[] visited, int start, int k) { if (k == 0) return true; if (curSum > eachSum) return false; if (curSum == eachSum) return helper(nums, eachSum, 0, visited, 0, k - 1); for (int i = start; i < nums.length; i++) { if (visited[i] == 1 || (i > start && nums[i] == nums[i - 1] && visited[i - 1] == 0)) continue; visited[i] = 1; if (helper(nums, eachSum, curSum + nums[i], visited, i + 1, k)) return true; visited[i] = 0; } return false; }
三、Matchsticks to Square
问题描述:
问题求解:
k = 4.
public boolean makesquare(int[] nums) { if (nums.length == 0) return false; int sum = 0; for (int num : nums) sum += num; if (sum % 4 != 0) return false; int eachSum = sum / 4; Arrays.sort(nums); return helper(nums, eachSum, 0, new int[nums.length], 0, 4); } private boolean helper(int[] nums, int eachSum, int curSum, int[] visited, int start, int k) { if (k == 0) return true; if (curSum > eachSum) return false; if (curSum == eachSum) return helper(nums, eachSum, 0, visited, 0, k - 1); for (int i = start; i < nums.length; i++) { if (visited[i] == 1 || (i > 0 && nums[i] == nums[i - 1] && visited[i - 1] == 0)) continue; visited[i] = 1; if (helper(nums, eachSum, curSum + nums[i], visited, i + 1, k)) return true; visited[i] = 0; } return false; }