Day 43 | 1049. Last Stone Weight II | 494. Target Sum | 474. Ones and Zeroes

原创已于 2023-08-16 23:22:02 修改 · 738 阅读

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于 2023-06-12 17:48:18 首次发布

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本文介绍了连续多日的编程题目，涉及二分查找算法、数组的处理（如寻找元素位置、删除元素），链表的操作（如反转、删除节点），树的遍历（前序、中序、后序）以及动态规划解题方法（如目标和、01背包问题）。这些题目涵盖了数据结构和算法的基础知识。

1049. Last Stone Weight II

Question Link

class Solution {
    public int lastStoneWeightII(int[] stones) {
        if(stones == null || stones.length == 0)
            return 0;
        int sum = 0;
        for(int stone : stones)
            sum += stone;

        int target = sum / 2;
        int[] dp = new int[target+1];
        for(int i = 0; i < stones.length; i++){
            for(int j = target; j >= stones[i]; j--)
                dp[j] = Math.max(dp[j], dp[j - stones[i]] + stones[i]);
        }
        return sum - 2 * dp[target];
    }
}

Divide the stones into two piles of equal weight as much as possible. The stone remaining after the collision is the smallest.
For instance, when input [2,4,1,1], target = (2 + 4 + 1 + 1)/2 = 4, status of 2D array:

	1	2	3	4
stones[0]	0	2	2	2
stones[1]	0	2	2	4
stones[2]	1	2	3	4
stones[3]	1	2	3	4

494. Target Sum

Question Link

class Solution {
    public int findTargetSumWays(int[] nums, int target) {
        int sum = 0;
        for(int num : nums)
            sum += num;
        if((sum + target)%2 != 0) return 0;
        if(target < 0 && -target > sum) return 0;

        int bagSize = (sum+target) / 2;
        int[] dp = new int[bagSize + 1];
        dp[0] = 1;

        for(int i = 0; i < nums.length; i++){
            for(int j = bagSize; j >= nums[i]; j--){
                dp[j] = dp[j] + dp[j - nums[i]];
            }
        }
        return dp[bagSize];
    }
}

Divide the nums in to an addition set left and a subtraction set right, then we can make the following derivation:
- left + right = sum
- left - right = target
- left - sum + left = target
- left = (target + sum) / 2
If sum + target can not be divided exactly by 2, which means that is problem has no solution.
Status change of 2D array：

	0	1	2	3	4
nums[0]	1	1	0	0	0
nums[1]	1	2	1	0	0
nums[2]	1	3	3	1	0
nums[3]	1	4	6	4	1
nums[4]	1	5	10	10	0

474. Ones and Zeroes

Question Link

class Solution {
    public int findMaxForm(String[] strs, int m, int n) {
        int[][] dp = new int[m+1][n+1];
        int oneNum, zeroNum;
        for(String str : strs){
            oneNum = 0;
            zeroNum = 0;
            for(char ch : str.toCharArray()){
                if(ch == '0')
                    zeroNum++;
                else
                    oneNum++;
            }
            // Reverse order traversal
            for(int i = m; i >= zeroNum; i--){
                for(int j = n; j >= oneNum ; j--)
                    dp[i][j] = Math.max(dp[i][j], dp[i - zeroNum][j - oneNum] + 1);
            }
        }
        return dp[m][n];
    }
}

dp[i][j]：The length of the largest subset of strs with at most i number of 0 and j number of 1
- Recursion formula: dp[i][j] = max(dp[i][j], dp[i - zeroNum][j - oneNum] + 1).
- Compare to the previous recursion formula, zeroNum and oneNum of the string are equivalent to the weight[i], the number of strings themselves corresponds to the value[i].
The elements of strs are the items in the bag, m and n correspond to the bag capacity.