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LeetCode Weekly Contest 38解题思路

版权声明:本文为博主原创文章,未经博主允许不得转载。 https://blog.csdn.net/u014688145/article/details/73718744

LeetCode Weekly Contest 38解题思路

详细代码可以fork下Github上leetcode项目,不定期更新。

赛题

本次周赛主要分为以下4道题:

Leetcode 628. Maximum Product of Three Numbers

Problem:

Given an integer array, find three numbers whose product is maximum and output the maximum product.

Example 1:

Input: [1,2,3] Output: 6

Example 2:

Input: [1,2,3,4] Output: 24

Note:

  • The length of the given array will be in range [3,104] and all elements are in the range [-1000, 1000].
  • Multiplication of any three numbers in the input won’t exceed the range of 32-bit signed integer.

求三个乘积最大,注意下负数的情况,两负*一正的情况也需要考虑。

代码如下:

    public int maximumProduct(int[] nums) {
        Arrays.sort(nums);
        int max = Integer.MIN_VALUE;
        int n = nums.length;
        max = Math.max(max, nums[0] * nums[1] * nums[2]);
        max = Math.max(max, nums[n-1] * nums[n-2] * nums[n-3]);
        max = Math.max(max, nums[0] * nums[1] * nums[n-1]);
        return max;
    }

Leetcode 630. Course Schedule III

Problem:

There are n different online courses numbered from 1 to n. Each course has some duration(course length) t and closed on dth day. A course should be taken continuously for t days and must be finished before or on the dth day. You will start at the 1st day. Given n online courses represented by pairs (t,d), your task is to find the maximal number of courses that can be taken.

Example:

Input: [[100, 200], [200, 1300], [1000, 1250], [2000, 3200]] Output: 3 Explanation: There’re totally 4 courses, but you can take 3 courses at most: First, take the 1st course, it costs 100 days so you will finish it on the 100th day, and ready to take the next course on the 101st day. Second, take the 3rd course, it costs 1000 days so you will finish it on the 1100th day, and ready to take the next course on the 1101st day. Third, take the 2nd course, it costs 200 days so you will finish it on the 1300th day. The 4th course cannot be taken now, since you will finish it on the 3300th day, which exceeds the closed date.

Note:

  • The integer 1 <= d, t, n <= 10,000.
  • You can’t take two courses simultaneously.

这道题说难还真有点难度,说简单也还挺简单的,只要有一个点get到了,通过归简技术就能解决了,也让我认识到归简的强大。

一些直观的认识,deadline最长的course应该放在最后处理,所以可以根据deadline从小大到排序,关键是如何得到正确的上课数呢?数学归纳法,能够证明,而我们只需要一种最优策略即可。

策略选择:

已知给定了n-1门课的deadline后,知道了这n-1门课中的最大上课数,根据当前信息选择是否加入第n门课。

当前信息是什么呢?还是一个直观的选择,既然在前n-1门课中,可以维护最大的上课数,为了让后续选择第n门课的成功率最高,选择n-1门课后,当前上的天数应该是最小的。

OKAY,有了这些就可以设计我们的代码了,如果成功加入第n门课,说明最大上课数变动了,所以只要更新now即可,而当第n门课未加入成功,则需要根据这第n门课求出最小的now。如下:

[6,7] [2,10],[4,11]

n = 2
now = 6 + 2 = 8
n = 3
发现加入第三门课时,8 + 4 > 11 说明最大还是2门课
但此时now = ?
如何让now最小?

这三门课中,删除持续天数最多的那门课
now = 6 + 2 + 4 - 6 = 6

所以需要用一个优先队列来维护一个最大持续天数,不断更新max相同情况下的now

证明:
now{n} 表示在集合n中,当前now的最小值,{}表示集中,当然这是我自己的表达方式
deadline[n] 表示在集合n中,当前第n门课的最小值

所以根据当前n-1门课的信息,可以得到:
a. now{n-1} <= deadline[n-1] <= deadline[n]
(deadline[n-1] <= deadline[n] 因为数组已经根据deadline排序)
加入第n课失败的情况下,我们有:
b. now{n-1} + courses[n][0] > courses[n][1]

现在需要求得now{n},假设:
now{n} = now{n-1} + courses[n][0]
为了最小化now,可以直接删除n门课中最大的持续天数:
c. now{n} = now{n-1} + courses[n][0] - courses[j][0]
需要证明 now{n} <= deadline[n],这样最大的上课数才能位置不变

c1. n == j
now{n} = now{n-1} <= deadline[n-1] <= deadline[n] 符合

c2. n != j
now{n} - now{n-1} = courses[n][0] - courses[j][0]

因为:
courses[j][0] > courses[n][0]

所以:
now{n} < now{n-1} <= deadline[n-1] <= deadline[n] 符合

综上,可以挑最大的courses[j][0]更新now,且最大max不变。

代码如下:

public int scheduleCourse(int[][] courses) {
        Arrays.sort(courses, (a, b) -> a[1] - b[1]);
        Queue<Integer> q = new PriorityQueue<>();
        int n = courses.length;
        int now = 0;
        for (int i = 0; i < n; ++i) {
            now += courses[i][0];
            q.offer(-courses[i][0]);
            while (now > courses[i][1]) {
                now += q.poll();
            }
        }
        return q.size();
    }

Leetcode 629. K Inverse Pairs Array

Problem:

Given two integers n and k, find how many different arrays consist of numbers from 1 to n such that there are exactly k inverse pairs. We define an inverse pair as following: For ith and jth element in the array, if i < j and a[i] > a[j] then it’s an inverse pair; Otherwise, it’s not. Since the answer may very large, the answer should be modulo 109 + 7.

Example 1:

Input: n = 3, k = 0 Output: 1 Explanation: Only the array [1,2,3] which consists of numbers from 1 to 3 has exactly 0 inverse pair.

Example 2:

Input: n = 3, k = 1 Output: 2 Explanation: The array [1,3,2] and [2,1,3] have exactly 1 inverse pair.

Note:

  • The integer n is in the range [1, 1000] and k is in the range [0, 1000].

一道DP题,此题考的是reverse的性质,就一个核心思想,如何根据n-1得到第n个数的所有组合。

看图:

上图说的当n = 5,k = 4时,根据n = 4得到的组合数。观察n = 5的表格你会发现,咱遍历的就是5的所有位置而已。这跟reverse的性质有关,因为添加的5比所有的1-4大,所一旦位置确定,新增的逆序对就确定,如在k = 1的情况,新增的逆序对个数为3,所以:

dp[5][4] = dp[5][4] + dp[4][1]; if pos所在的位置的逆序对个数为3

当pos = 3时,新增的逆序对为2,所以有:
dp[5][4] = dp[5][4] + dp[4][2]

代码如下:

    static int MOD = 1000000000 + 7;
    public int kInversePairs(int n, int k) {
        int[][] dp = new int[n + 1][k + 1];
        for (int i = 1; i <= n; ++i)
            dp[i][0] = 1;
        for (int i = 2; i <= n; ++i) {
            for (int j = 1; j <= k; ++j) {
                for (int l = Math.min(i - 1, j); l >= 0; --l) {
                    dp[i][j] = (dp[i][j] + dp[i - 1][j - l]) % MOD;
                }
            }
        }
        return dp[n][k];
    }

TLE,时间复杂度为O(n2⋅k)O(n^2 \cdot k),基本思想有了,就看怎么优化了。可以参考Leetcode官方提供的教程https://leetcode.com/articles/k-inverse-pairs-array/,其中在解法4中,用到的就是上述更新过程,观察下移动的性质,你会发现它实际就是一种滑动窗口的更新模式,所以解法五,利用了dp[i][j-1]的值,在其基础上加上dp[i-1][j]再减去dp[i-1][j-1]即可。

代码如下:

static int MOD = 1000000000 + 7;
    public int kInversePairs(int n, int k) {
        int[][] dp = new int[n + 1][k + 1];
        for (int i = 1; i <= n; ++i) dp[i][0] = 1;
        for (int i = 2; i <= n; ++i) {
            for (int j = 1; j <= k; ++j) {
                int val = (dp[i-1][j] - ((j - i < 0) ? 0 : dp[i-1][j-i]) + MOD) % MOD;
                dp[i][j] = (dp[i][j-1] + val) % MOD;
            }
        }
        return dp[n][k];
    }

Leetcode 631. Design Excel Sum Formula

Problem:

Your task is to design the basic function of Excel and implement the function of sum formula. Specifically, you need to implement the following functions: Excel(int H, char W): This is the constructor. The inputs represents the height and width of the Excel form. H is a positive integer, range from 1 to 26. It represents the height. W is a character range from ‘A’ to ‘Z’. It represents that the width is the number of characters from ‘A’ to W. The Excel form content is represented by a height * width 2D integer array C, it should be initialized to zero. You should assume that the first row of C starts from 1, and the first column of C starts from ‘A’. void Set(int row, char column, int val): Change the value at C(row, column) to be val. int Get(int row, char column): Return the value at C(row, column). int Sum(int row, char column, List of Strings : numbers): This function calculate and set the value at C(row, column), where the value should be the sum of cells represented by numbers. This function return the sum result at C(row, column). This sum formula should exist until this cell is overlapped by another value or another sum formula. numbers is a list of strings that each string represent a cell or a range of cells. If the string represent a single cell, then it has the following format : ColRow. For example, “F7” represents the cell at (7, F). If the string represent a range of cells, then it has the following format : ColRow1:ColRow2. The range will always be a rectangle, and ColRow1 represent the position of the top-left cell, and ColRow2 represents the position of the bottom-right cell.

Example 1:

Excel(3,”C”); // construct a 3*3 2D array with all zero. // A B C // 1 0 0 0 // 2 0 0 0 // 3 0 0 0 Set(1, “A”, 2); // set C(1,”A”) to be 2. // A B C // 1 2 0 0 // 2 0 0 0 // 3 0 0 0 Sum(3, “C”, [“A1”, “A1:B2”]); // set C(3,”C”) to be the sum of value at C(1,”A”) and the values sum of the rectangle range whose top-left cell is C(1,”A”) and bottom-right cell is C(2,”B”). Return 4. // A B C // 1 2 0 0 // 2 0 0 0 // 3 0 0 4 Set(2, “B”, 2); // set C(2,”B”) to be 2. Note C(3, “C”) should also be changed. // A B C // 1 2 0 0 // 2 0 2 0 // 3 0 0 6

Note:

  • You could assume that there won’t be any circular sum reference. For example, A1 = sum(B1) and B1 = sum(A1).
  • The test cases are using double-quotes to represent a character.
  • Please remember to RESET your class variables declared in class Excel, as static/class variables are persisted across multiple test cases. Please see here for more details.

好吧,重在理解题意,set的时候需要把当前的公式清零,而在get的时候,map中存在它的求和公式,则先更新,再返回。接着在sum中记录所有的公式即可。

代码如下:

public class Excel {

    int[][] table;
    Map<String, String[]> mem = new HashMap<>();

    public Excel(int H, char W) {
        table = new int[H][W - 'A' + 1];
    }

    public void set(int r, char c, int v) {
        table[r-1][c - 'A'] = v;
        if (mem.containsKey(""+c+r)) mem.remove(""+c+r);
    }

    public int get(int r, char c) {
        if(mem.containsKey(""+r+c)) sum(r,c,mem.get(""+r+c));
        return table[r - 1][c - 'A'];
    }

    public int sum(int r, char c, String[] strs) {
        mem.put(""+c+r,strs);
        int sum = 0;
        for (int i = 0; i < strs.length; ++i){
            String cell = strs[i];
            if (cell.contains(":")){
                String[] cells = cell.split(":");
                char topC = cells[0].charAt(0);
                char botC = cells[1].charAt(0);
                int topR = 0;
                for (int j = 1; j < cells[0].length(); ++j) topR = topR * 10 + cells[0].charAt(j) - '0';
                int botR = 0;
                for (int j = 1; j < cells[1].length(); ++j) botR = botR * 10 + cells[1].charAt(j) - '0';
                for (int k = topR; k <= botR; ++k){
                    for (char l = topC; l <= botC; ++l){
                        sum += get(k, l);
                    }
                }
            }
            else{
                char col = cell.charAt(0);
                int row = 0;
                for (int j = 1; j < cell.length(); ++j) row = row * 10 + cell.charAt(j) - '0';
                sum += get(row, col);
            }
        }
        return table[r-1][c - 'A'] = sum;
    }
}

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