LeetCode 0-1 Knapsack 背包问题&相关题目

关于我的 Leetcode 题目解答，代码前往 Github：https://github.com/chenxiangcyr/leetcode-answers

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Given weights and values of n items, put these items in a knapsack of capacity W to get the maximum total value in the knapsack.
In other words, given two integer arrays val[0..n-1] and wt[0..n-1] which represent values and weights associated with n items respectively.
Also given an integer W which represents knapsack capacity, find out the maximum value subset of val[] such that sum of the weights of this subset is smaller than or equal to W. You cannot break an item, either pick the complete item, or don’t pick it (0-1 property).

代码如下，包括递归和非递归算法。其时间复杂度均为 O(n * W)

class Knapsack {

    // Returns the maximum value that can be put in a knapsack of capacity W
    static int knapSackRecursive(int W, int wt[], int val[], int n) {
        // Base Case
        if (n == 0 || W == 0)
            return 0;

        // If weight of the nth item is more than Knapsack capacity W, then
        // this item cannot be included in the optimal solution
        if (wt[n - 1] > W)
            return knapSackRecursive(W, wt, val, n - 1);

        // Return the maximum of two cases:
        // (1) nth item included
        // (2) not included
        else
            return Math.max(
                    val[n - 1]
                            + knapSackRecursive(W - wt[n - 1], wt, val, n - 1),
                    knapSackRecursive(W, wt, val, n - 1));
    }

    // Returns the maximum value that can be put in a knapsack of capacity W
    static int knapSackIterative(int W, int wt[], int val[], int n) {

        // dp[i][j] stores the maximum value when there are i element and the
        // knapstack is j
        int[][] dp = new int[n + 1][W + 1];
        dp[0][0] = 0;

        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= W; j++) {
                if (wt[i - 1] > j) {
                    dp[i][j] = dp[i - 1][j];
                } else {
                    /*
                     * case1: 带上第i个元素，获得收益val[i] case2: 不带上第i个元素
                     */
                    int val1 = val[i - 1] + dp[i - 1][j - wt[i - 1]];
                    int val2 = dp[i - 1][j];

                    dp[i][j] = Math.max(val1, val2);
                }
            }
        }

        return dp[n][W];
    }

    // Driver program to test above function
    public static void main(String args[]) {
        int val[] = new int[] { 60, 100, 120 };
        int wt[] = new int[] { 10, 20, 30 };
        int W = 51;
        int n = val.length;
        System.out.println(knapSackIterative(W, wt, val, n));
    }
}

LeetCode题目：518. Coin Change 2
You are given coins of different denominations and a total amount of money. Write a function to compute the number of combinations that make up that amount. You may assume that you have infinite number of each kind of coin.

class Solution {
    public int change(int amount, int[] coins) {
        /*
        dp[i][j] 表示使用前i种硬币组成j块钱的组合数
        */
        int[][] dp = new int[coins.length + 1][amount + 1];
        dp[0][0] = 1;
        
        for(int i = 1; i <= coins.length; i++) {
            dp[i][0] = 1;
            
            for(int j = 1; j <= amount; j++) {
                
                int coin = coins[i - 1];
                /*
                case1: 使用第i种硬币
                case2: 不使用第i种硬币
                */
                dp[i][j] = dp[i - 1][j] + (j >= coin ? dp[i][j - coin] : 0);
                
            }
        }
        
        return dp[coins.length][amount];
    }
}

LeetCode题目：416. Partition Equal Subset Sum
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].

class Solution {
    public boolean canPartition(int[] nums) {
        
        if(nums == null || nums.length < 2) return false;
        
        int sum = 0;
        for(int i : nums) {
            sum = sum + i;
        }
        
        // odd
        if(sum % 2 == 1) return false;
        
        sum = sum /2;
        
        // 0-1背包问题
        int n = nums.length;
        
        // dp[i][j] 表示使用前 i 个元素求和得到 j 是否可能
        boolean[][] dp = new boolean[n + 1][sum + 1];
        
        for (int i = 0; i < dp.length; i++) {
            Arrays.fill(dp[i], false);
        }

        dp[0][0] = true;

        // 第一列
        for (int i = 1; i < n + 1; i++) {
            dp[i][0] = true;
        }
        
        // 第一行
        for (int j = 1; j < sum + 1; j++) {
            dp[0][j] = false;
        }

        for (int i = 1; i < n + 1; i++) {
            for (int j = 1; j < sum + 1; j++) {
                
                if (j >= nums[i - 1]) {
                    dp[i][j] = (dp[i - 1][j] || dp[i - 1][j - nums[i - 1]]);
                }
                
                else {
                    dp[i][j] = dp[i - 1][j];
                }
            }
        }

        return dp[n][sum];
    }
}

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引用：
Dynamic Programming | Set 10 ( 0-1 Knapsack Problem)