【重点】【背包】【DP】416.分割等和子集

题目

法1：dp

基础解法，必须掌握！！！
本质是『0-1背包』问题，背包问题大总结

Python

class Solution:
    def canPartition(self, nums: List[int]) -> bool:
        n = len(nums)
        if sum(nums) % 2 == 1 or max(nums) > sum(nums) / 2:
            return False
        target = sum(nums) // 2
        dp = [[False] * (target + 1) for _ in range(n)] # dp[i][j]表示{[nums[0], nums[i]}之间的数字能否组成和j
        for i in range(n):
            dp[i][nums[i]] = True
            dp[i][0] =  True
        
        for i in range(1, n):
            for j in range(1, target+1):
                if nums[i] > j:
                    dp[i][j] = dp[i-1][j] # 不选择i
                else:
                    dp[i][j] = dp[i-1][j] or dp[i-1][j-nums[i]] # 不选 or 选择 
        
        return dp[n-1][target]

Java

class Solution {
    public boolean canPartition(int[] nums) {
        if (nums.length < 2) {
            return false;
        }
        int sum = 0, maxVal = 0;
        for (int i : nums) {
            sum += i;
            maxVal = i > maxVal ? i : maxVal;
        }
        if (sum % 2 == 1 || maxVal > sum / 2) {
            return false;
        }

        int n = nums.length, target = sum / 2;
        boolean[][] dp = new boolean[n][target + 1]; // dp[i][j]表示能否从[0, i]组合sum为j
        for (int i = 0; i < nums.length; ++i) {
            dp[i][nums[i]] = true;
            dp[i][0] = true;
        }

        for (int i = 1; i < nums.length; ++i) {
            for (int j = 1; j <= target; ++j) {
                if (j < nums[i]) {
                    dp[i][j] = dp[i - 1][j];
                } else {
                    dp[i][j] = dp[i - 1][j] || dp[i - 1][j - nums[i]]; // 取nums[i] or not
                }
            }
        }

        return dp[n - 1][target];
    }
}