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.
本题要找是否给定的正数数组是否能被分成两部分,使两部分的和相等。也就是说,需要从物品(数组)中,找到价值(和)为数组总和的一半的物品,放进背包。
如果能够找到,返回true,否则false。
class Solution { public boolean canPartition(int[] nums) { // 数组求和 int sum = 0; for(int num : nums) sum += num; // 如果和是奇数,说明价值不可能正好为和的一半,结果为false if(sum%2!=0) return false; int[] dp = new int[sum]; return canPartition(nums, 0, sum/2, 0, dp); } public boolean canPartition(int[] nums, int sum, int target, int i, int[] dp) { // 当前和为目标 if(sum==target) return true; // i超过界限或者和比目标大 if(i>=nums.length || sum>target) return false; // 之前储存了结果,不再计算一遍,直接返回 // 用 int[] dp 而不用 boolean[] dp 是为了区分已计算的和未计算的 if(dp[sum]!=0) return dp[sum]==1 ? true : false; boolean res = canPartition(nums, sum+nums[i], target, i+1, dp) || canPartition(nums, sum, target, i+1, dp); // 当前结果存到dp中 dp[sum] = res ? 1 : -1; return res; } }