416.md

#416. Partition Equal Subset Sum 题目链接

public class Solution {
    /*
        转化为01背包问题
     */
    public boolean canPartition(int[] nums) {
        // 首先判断一些特殊情况
        int sum = 0;
        for (int i = 0; i < nums.length; i++) {
            sum += nums[i];
        }
        if (sum % 2 == 1) {
            return false;
        }
        int target = sum / 2;
        for (int i = 0; i < nums.length; i++) {
            if (nums[i] > target) {
                return false;
            } else if (nums[i] == target) {
                return true;
            }
        }

        // 如果仍无法判断，转化为01背包问题
        int len = nums.length;
        boolean[][] dp = new boolean[len + 1][target + 1];
        for (int i = 0; i < dp.length; i++) {
            Arrays.fill(dp[i], false);
        }
        for (int i = 0; i < len + 1; i++) {
            dp[i][0] = true;
        }
        for (int i = 1; i < len + 1; i++) {
            for (int j = 1; j < target + 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[len][target];
    }
}

另外一个降低了空间复杂度的01背包方案

public class Solution {
    /*
        转化为01背包问题
     */
    public boolean canPartition(int[] nums) {
        // 首先判断一些特殊情况
        int sum = 0;
        for (int i = 0; i < nums.length; i++) {
            sum += nums[i];
        }
        if (sum % 2 == 1) {
            return false;
        }
        int target = sum / 2;
        for (int i = 0; i < nums.length; i++) {
            if (nums[i] > target) {
                return false;
            } else if (nums[i] == target) {
                return true;
            }
        }

        // 如果仍无法判断，转化为01背包问题
        int len = nums.length;
        boolean[] dp = new boolean[target + 1];
        Arrays.fill(dp, false);
        dp[0] = true;
        for (int i = 1; i < len + 1; i++) {
            // 注意这里的方向是从后向前
            for (int j = target; j > 0; j--) {
                if (j >= nums[i - 1]) {
                    dp[j] = dp[j] || dp[j - nums[i - 1]];
                }
            }
        }
        return dp[target];
    }
}