算法一

Write a function:

int solution(vector<int>&A);

that, given an array A of length N, returns as an integer the minimum number of moves needed to transform a zero-filled array into A.

Examples:

1. Given A = [2, 1, 3], the function should return 4. For example, the following sequence of moves leads to the solution: [0, 0, 0] → [1, 1, 1] → [2, 1, 1]→ [2, 1, 2] → [2, 1, 3].

2. Given A = [2, 2, 0, 0, 1], the function should return 3. The following sequence of moves leads to the solution: [0, 0, 0, 0, 0] → [1, 1, 0, 0, 0] → [2, 2, 0, 0, 0] → [2, 2, 0, 0, 1].

3. Given A = [5, 4, 2, 4, 1], the function should return 7. One possible optimal sequence of moves is: [0, 0, 0, 0, 0] → [1, 1, 1, 1, 1] → [2, 2, 2, 2, 1] → [3, 3, 2, 2, 1] → [4, 4, 2, 2, 1] → [5, 4, 2, 2, 1] → [5, 4, 2, 3, 1] → [5, 4, 2, 4, 1].

Write an efficient algorithm for the following assumptions:

N is an integer within the range [1..100,000]; each element of array A is an integer within therange [0..1,000,000,000]:

we guarantee that the answer will not exceed1,000,000,000.

上面这道题大概就是将一个全0数组变成到目标数组需要多少步骤。每次只能对连续的数字进行增加1的操作。

# vector<int>& A
def solution(A) {int n = len(A);if (n == 0) return 0;# Variable to store the total number of moves requiredint moves = A[0];  # We need at least A[0] moves to reach the first element's valuefor (int i = 1; i < n; ++i) {# If current element is greater than the previous one, # we need extra moves to reach itif (A[i] > A[i - 1]) {moves += (A[i] - A[i - 1]);}}return moves;
}

算法二

There is an array, named `digits`, consisting of N digits. Choose at most three digits (not necessarily adjacent) and merge them into a new integer without changing the order of the digits. What is the biggest number that can be obtained this way?

Write a function:

def solution(digits)

that, given an array of N digits, returns the biggest number that can be built.

Examples:
1. Given digits = [7, 2, 3, 3, 4, 9], the function
should return 749.
2. Given digits = [0, 0, 5, 7], the function should
return 57.

Assume that:
N is an integer within the range [3..50];
each element of array, named `digits`, is an
integer within the range [0..9].

In your solution, focus on correctness. The
performance of your solution will not be the focus of
the assessment.

算法是要从一个数组里面选出三个数用来组成一个新的三位数，这三个数字的相对位置不能改变，求能够组成的最大的数字。

算法思路就是先从0~n-2之前挑出最大的数字，然后从上一个数字的索引往后直到n-1之前挑出第二个最大的数字，再从上一个数字的索引到n挑出第三大的数字即可。