LintCode 749: John's backyard garden (完全背包题变种)

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于 2019-12-03 16:48:08 发布

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分类专栏： algorithm-design 文章标签： LintCode

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本文探讨了一个有趣的算法问题，即如何使用两种不同高度的砖块（3dm和7dm）来构建任意高度的围墙。通过一个完全背包变种算法的实现，我们能够判断是否可以使用这些砖块构建出特定高度的围墙。本文提供了详细的算法思路及代码实现，为读者展示了如何解决此类问题。

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John’s backyard garden

John wants to build a back garden on the empty space behind his home. There are two kinds of bricks now, one is 3 dm high and the other is 7 dm high. John wants to enclose a high x dm wall. If John can do this, output YES, otherwise NO.

Example
Example 1:

Input : x = 10
Output : “YES”
Explanation :
x = 3 + 7:That is, you need one batch of 3 dm height bricks and one batch of 7 dm height bricks.
Example 2:

Input : x = 5
Output : “NO”
Explanation:
John can not enclose a high 5 dm wall with 3 dm height bricks and 7 dm height bricks.
Example 3:

Input : x = 13
Output : “YES”
Explanation :
x = 2 * 3 + 7:That is, you need two batch of 3 dm height bricks and one batch of 7 dm height bricks.
Notice
X is an integer, and it’s range is [3, 1000].

解法1：完全背包变种。
代码如下：

class Solution {
public:
    /**
     * @param x: the wall's height
     * @return: YES or NO
     */
    string isBuild(int x) {
        string result;
        vector<int> A = {3, 7};
        vector<int> dp(x + 1, 0);
        dp[0] = 1;
        for (int i = 0; i <= 1; ++i) {
            for (int j = A[i]; j <= x; ++j) {
                dp[j] = dp[j] || dp[j - A[i]];
            }    
        }
        
        if (dp[x]) return "YES";
        else return "NO";
    }
};