[LeetCode][416]partition-equal-subset-sum
Content
Given an integer array nums
, return true
if you can partition the array into two subsets such that the sum of the elements in both subsets is equal or false
otherwise.
Example 1:
Input: nums = [1,5,11,5] Output: true Explanation: The array can be partitioned as [1, 5, 5] and [11].
Example 2:
Input: nums = [1,2,3,5] Output: false Explanation: The array cannot be partitioned into equal sum subsets.
Constraints:
1 <= nums.length <= 200
1 <= nums[i] <= 100
Related Topics
Solution
1.动态规划 + 01背包
Java
class Solution {
public boolean canPartition(int[] nums) {
// 1 <= nums.length <= 200
// 1 <= nums[i] <= 100
int sum = 0;
for (int num : nums) {
sum += num;
}
if ((sum & 1) == 1) {
return false;
}
int half = sum >> 1;
int n = nums.length;
// dp[i][j]表示前i个数字中任意个数字累加的最大值,上限为j
int[][] dp = new int[n + 1][half + 1];
for (int i = 1; i <= n; i++) {
int num = nums[i - 1];
if (num <= half) {
dp[i][half] = num + dp[i - 1][half - num];
if (dp[i][half] == half) {
return true;
}
}
for (int j = half - 1; j >= 0; j--) {
// 不累加当前数字
dp[i][j] = dp[i - 1][j];
// 累加当前数字
if (num <= j) {
dp[i][j] = Math.max(dp[i][j], num + dp[i - 1][j - num]);
}
}
}
return false;
}
}
2.动态规划(空间优化)
Java
class Solution {
public boolean canPartition(int[] nums) {
// 1 <= nums.length <= 200
// 1 <= nums[i] <= 100
int sum = 0;
for (int num : nums) {
sum += num;
}
if ((sum & 1) == 1) {
return false;
}
int half = sum >> 1;
int n = nums.length;
// dp[j]表示前i个数字中任意个数字累加的最大值,上限为j
int[] dp = new int[half + 1];
for (int i = 1; i <= n; i++) {
int num = nums[i - 1];
if (num <= half) {
dp[half] = num + dp[half - num];
if (dp[half] == half) {
return true;
}
}
for (int j = half - 1; j > 0; j--) {
// 累加当前数字
if (num <= j) {
dp[j] = Math.max(dp[j], num + dp[j - num]);
}
}
}
return false;
}
}