要死要死,第一题竟然错误8次,心态崩了呀,自己没有考虑清楚,STL用的也不是很熟,一直犯错。
第二题也是在室友的帮助下完成的,心态崩了。
970. Powerful Integers
Given two non-negative integers x
and y
, an integer is powerful if it is equal to x^i + y^j
for some integers i >= 0
and j >= 0
.
Return a list of all powerful integers that have value less than or equal to bound
.
You may return the answer in any order. In your answer, each value should occur at most once.
Example 1:
Input: x = 2, y = 3, bound = 10 Output: [2,3,4,5,7,9,10] Explanation: 2 = 2^0 + 3^0 3 = 2^1 + 3^0 4 = 2^0 + 3^1 5 = 2^1 + 3^1 7 = 2^2 + 3^1 9 = 2^3 + 3^0 10 = 2^0 + 3^2
Example 2:
Input: x = 3, y = 5, bound = 15 Output: [2,4,6,8,10,14]
Note:
1 <= x <= 100
1 <= y <= 100
0 <= bound <= 10^6
题目意思很清楚就不解释了。说下两个很坑的数据吧,输出结果可以任意顺序, 第一个:输入 1 1 1 输出 [1] 第二个 1 2 100 输出 [2,3,5,9,17,33,65]
class Solution { public: vector<int> powerfulIntegers(int x, int y, int bound) { vector<int> ans; if( x < y ) swap(x, y); int px = 1, py = 1; while( px <= bound ) { py = 1; while( px + py <= bound ) { ans.push_back(px+py); py = py*y; if( y == 1 ) break; } px = px * x; if( x == 1 ) break; } sort(ans.begin(), ans.end()); ans.erase(unique(ans.begin(), ans.end()), ans.end()); return ans; } };
969. Pancake Sorting
Given an array A
, we can perform a pancake flip: We choose some positive integer k <= A.length
, then reverse the order of the first kelements of A
. We want to perform zero or more pancake flips (doing them one after another in succession) to sort the array A
.
Return the k-values corresponding to a sequence of pancake flips that sort A
. Any valid answer that sorts the array within 10 * A.length
flips will be judged as correct.
Example 1:
Input: [3,2,4,1] Output: [4,2,4,3] Explanation: We perform 4 pancake flips, with k values 4, 2, 4, and 3. Starting state: A = [3, 2, 4, 1] After 1st flip (k=4): A = [1, 4, 2, 3] After 2nd flip (k=2): A = [4, 1, 2, 3] After 3rd flip (k=4): A = [3, 2, 1, 4] After 4th flip (k=3): A = [1, 2, 3, 4], which is sorted.
Example 2:
Input: [1,2,3] Output: [] Explanation: The input is already sorted, so there is no need to flip anything. Note that other answers, such as [3, 3], would also be accepted.
Note:
1 <= A.length <= 100
A[i]
is a permutation of[1, 2, ..., A.length]
题意:规定了翻转规则是:选择一个数k将数组前面的k个数字移到数组后面。现在问你给你一个打乱顺序的数组A,你要尽量多少次这样的翻转才能将其变成一个有序的数组(1-n排列)。
思路:无论怎么翻转,最后的结果肯定是A[i]=i+1,所以直接去数组中找对应的数字然后直接按照规则翻转就好。
class Solution { public: vector<int> pancakeSort(vector<int>& A) { vector<int> ans; int n = A.size(); for (int i = n; i > 0; i--) { int index = find(A.begin(), A.end(), i) - A.begin(); ans.push_back(index + 1); reverse(A.begin(), A.begin() + index + 1); ans.push_back(i); reverse(A.begin(), A.begin() + i); } return ans; } };
原文地址:https://www.cnblogs.com/Asimple/p/10228326.html