心得

題目詢問說共可以滿足幾個小孩所期望的蛋糕數，例如總共有三個小孩A, B, C，A希望獲得1片蛋糕；B希望獲得2片蛋糕；C希望獲得3片蛋糕，而總共有兩塊蛋糕Z, X，Z可以切成1片，X可以切成兩片，這樣的話答案是2，因為最多只能滿足兩個小孩的需求。依照這個思路去撰寫Code就容易得多了。

問題

Assume you are an awesome parent and want to give your children some cookies. But, you should give each child at most one cookie. Each child i has a greed factor g_i, which is the minimum size of a cookie that the child will be content with; and each cookie j has a size s_j. If s_j >= g_i, we can assign the cookie j to the child i, and the child i will be content. Your goal is to maximize the number of your content children and output the maximum number.

Note:
You may assume the greed factor is always positive.
You cannot assign more than one cookie to one child.

Example 1:

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

Output: 1

Explanation: You have 3 children and 2 cookies. The greed factors of 3 children are 1, 2, 3. 
And even though you have 2 cookies, since their size is both 1, you could only make the child whose greed factor is 1 content.
You need to output 1.

Example 2:

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

Output: 2

Explanation: You have 2 children and 3 cookies. The greed factors of 2 children are 1, 2. 
You have 3 cookies and their sizes are big enough to gratify all of the children, 
You need to output 2.

答案

public class Solution {
    public int FindContentChildren(int[] g, int[] s) {
        Array.Sort(g);
        Array.Sort(s);

        int ans = 0;

        for (int i = 0; ans < g.Length && i < s.Length; i++)
        {
            if (g[ans] <= s[i])
            {
                ans++;
            }
        }
        
        return ans;
    }
}

心得

簡單來說就是輸入數字x，找出x的因數並找出相乘會等於x的兩個數字且兩數字相減最接近0的組合，大的放前面小的放後面以int[]的型態回傳。

問題

For a web developer, it is very important to know how to design a web page’s size. So, given a specific rectangular web page’s area, your job by now is to design a rectangular web page, whose length L and width W satisfy the following requirements:

1. The area of the rectangular web page you designed must equal to the given target area.

2. The width W should not be larger than the length L, which means L >= W.

3. The difference between length L and width W should be as small as possible.

You need to output the length L and the width W of the web page you designed in sequence.

Example:

Input: 4
Output: [2, 2]
Explanation: The target area is 4, and all the possible ways to construct it are [1,4], [2,2], [4,1]. 
But according to requirement 2, [1,4] is illegal; according to requirement 3,  [4,1] is not optimal compared to [2,2]. So the length L is 2, and the width W is 2.

Note:

1. The given area won’t exceed 10,000,000 and is a positive integer
2. The web page’s width and length you designed must be positive integers.

答案

1. 我的笨方法

   public class Solution {
       public int[] ConstructRectangle(int area) {
           int num1 = Enumerable
               .Range(1, (int)Math.Sqrt(area))
               .Where(x => area % x == 0)
               .OrderBy(x => Math.Abs((area / x) - x))
               .FirstOrDefault();
           int num2 = area / num1;
           return num1 > num2 ? new int[] { num1, num2 } : new int[] { num2, num1 };
       }
   }

2. Top Solution的神方法

   public class Solution {
       public int[] ConstructRectangle(int area) {
           int w = (int)Math.Sqrt(area);
           while( area % w != 0){
               w--;
           }
           return new int[] { area / w, w };
       }
   }

    

心得

傳入一個二進位的陣列，找出1最多連續出現的次數。

問題

Given a binary array, find the maximum number of consecutive 1s in this array.

Example 1:

Input: [1,1,0,1,1,1]
Output: 3
Explanation: The first two digits or the last three digits are consecutive 1s.
    The maximum number of consecutive 1s is 3.

Note:

• The input array will only contain 0 and 1.
• The length of input array is a positive integer and will not exceed 10,000

答案

public class Solution {
    public int FindMaxConsecutiveOnes(int[] nums) {
        int maxcount = 0, count = 0;
        for (int i = 0; i < nums.Length; i++)
        {
            if (nums[i] == 1)
            {
                count++;
            }
            else
            {
                if (maxcount < count)
                {
                    maxcount = count;
                }
                count = 0;
            }
        }
        return maxcount < count ? count : maxcount;
    }
}

心得

跟 [C#][LeetCode][Easy] 283. Move Zeroes 有87% 像。

問題

Given an array and a value, remove all instances of that value in place and return the new length.

Do not allocate extra space for another array, you must do this in place with constant memory.

The order of elements can be changed. It doesn’t matter what you leave beyond the new length.

Example:
Given input array nums = [3,2,2,3], val = 3

Your function should return length = 2, with the first two elements of nums being 2.

答案

public class Solution {
    public int RemoveElement(int[] nums, int val) {
        int j = 0;
        for(int i = 0; i < nums.Length; i++){
            if(nums[i] != val){
                nums[j] = nums[i];
                j++;
            }
        }
        
        return j;
    }
}

心得

將數字陣列中為零的數字在不更動其他數字排序的情況下移至陣列最後方，這題要求必須使用原陣列不能new一個新的物件就比較麻煩一點點了。

問題

Given an array nums, write a function to move all 0‘s to the end of it while maintaining the relative order of the non-zero elements.

For example, given nums = [0, 1, 0, 3, 12], after calling your function, nums should be [1, 3, 12, 0, 0].

Note:

1. You must do this in-place without making a copy of the array.
2. Minimize the total number of operations.

答案

public class Solution {
    public void MoveZeroes(int[] nums) {
        int j = 0;
        for(int i = 0; i < nums.Length; i++){
            if(nums[i] != 0){
                nums[j] = nums[i];
                j++;
            }
        }
        
        for(int i = j; i < nums.Length; i++){
            nums[i] = 0;
        }
    }
}