你可不可以
成为我的main函数
做我此生必须有
且只能有一个的入口
——————————
我愿为自己加上private
在你的class中只有
你能调用
——————————————代 码 如 诗 。
sort(a, a+5); // 默认从小到大,int数组的排序
sort(v.begin(), v.end()); // vector数组排序
sort中使用的是快排和插排
stable_sort(v.begin(), v.end());
partial_sort(a, a+2, a+n); // 前2个数字将是正确顺序
partial_sort(v.begin(), v.begin() + 3, v.end()); // 对数组中元素部分排序，前3个将是正确顺序
nth_element(v.begin(), v.begin() + 6, v.end()); // 只有v[6]处放置的是正确的元素，其余不管～
sort(v.begin(), v.end(), greater<int>()); // 从大到小排，algorithm头文件
bool flag = is_sorted(a, a+n);
bool flag = is_sort(v.begin(), v.end());
On the first row, we write a 0. Now in every subsequent row, we look at the previous row and replace each occurrence of 0 with 01, and each occurrence of 1 with 10.
Given row N and index K, return the K-th indexed symbol in row N. (The values of K are 1-indexed.) (1 indexed).
Examples:
Input: N = 1, K = 1
Output: 0
Input: N = 2, K = 1
Output: 0
Input: N = 2, K = 2
Output: 1
Input: N = 4, K = 5
Output: 1
Explanation:
row 1: 0
row 2: 01
row 3: 0110
row 4: 01101001
Note:
N will be an integer in the range [1, 30].
K will be an integer in the range [1, 2^(N-1)].
题目大意：在第一行，我们写一个0.现在在后面的每一行中，我们看前一行，用01代替0出现的每一个，每一次出现1代表10。现在问第N行第K数字是什么数字
分析：用递归，已知N == 1的时候返回0，为了知道第N行第K个数字的值，只要知道它在第N-1行的第(K+1)/2个数字对应的值即可，因为0对应01，1对应10，那么如果K是奇数只需和原数字相同即可，如果K是偶数只需对原对应数字取反即可～
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class Solution { public: int kthGrammar(int N, int K) { if (N == 1) return 0; return (K % 2 == 0) ? !kthGrammar(N-1, (K+1)/2) : kthGrammar(N-1, (K+1)/2); } }; |
Given an array consisting of n integers, find the contiguous subarray of given length k that has the maximum average value. And you need to output the maximum average value.
Example 1:
Input: [1,12,-5,-6,50,3], k = 4
Output: 12.75
Explanation: Maximum average is (12-5-6+50)/4 = 51/4 = 12.75
Note:
1 <= k <= n <= 30,000.
Elements of the given array will be in the range [-10,000, 10,000].
题目大意：给定一个由n个整数组成的数组，找到具有最大平均值的给定长度k的连续子数组。你需要输出最大的平均值～
分析：sum[i]保存0～i个数字的和，每次将(sum[i]-sum[i-k])的最大值保存在ans中，然后返回ans*1.0/k
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class Solution { public: double findMaxAverage(vector<int>& nums, int k) { if (nums.size() == 0) return 0.0; vector<int> sum(nums.size()); sum[0] = nums[0]; for (int i = 1; i < k; i++) sum[i] = sum[i-1] + nums[i]; int ans = sum[k-1]; for (int i = k; i < nums.size(); i++) { sum[i] = sum[i-1] + nums[i]; ans = max(ans, sum[i] - sum[i-k]); } return ans * 1.0 / k; } }; |
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.
题目大意：给一个数组，找三个数字，使这三个数乘积最大，返回乘积的最大值～
分析：先对数组排序，要么是最后三个数字乘积，要么可能会有负数，则是前两个数的乘积*最后一个数字的乘积，取这个两个乘积结果的最大值即可～
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class Solution { public: int maximumProduct(vector<int>& nums) { sort(nums.begin(), nums.end()); int n = nums.size(); return max(nums[0] * nums[1] * nums[n-1], nums[n-3] * nums[n-2] * nums[n-1]); } }; |
Given an unsorted array of integers, find the length of longest continuous increasing subsequence (subarray).
Example 1:
Input: [1,3,5,4,7]
Output: 3
Explanation: The longest continuous increasing subsequence is [1,3,5], its length is 3.
Even though [1,3,5,7] is also an increasing subsequence, it is not a continuous one where 5 and 7 are separated by 4.
Example 2:
Input: [2,2,2,2,2]
Output: 1
Explanation: The longest continuous increasing subsequence is [2], its length is 1.
Note: Length of the array will not exceed 10,000.
题目大意：给一个乱序数组，返回这个数组中递增连续子序列中最长的长度～
分析：遍历从i = 1到结尾，每次比较nums[i-1]和nums[i]，如果是递增的就将temp++，否则temp=1，每次将temp最大值保存在ans中，最后返回ans的值～
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class Solution { public: int findLengthOfLCIS(vector<int>& nums) { int temp = 1, ans = 1; for (int i = 1; i < nums.size(); i++) { temp = (nums[i-1] < nums[i]) ? temp + 1 : 1; ans = max(ans, temp); } return nums.size() == 0 ? 0 : ans; } }; |
Given a non-negative integer c, your task is to decide whether there’re two integers a and b such that a2 + b2 = c.
Example 1:
Input: 5
Output: True
Explanation: 1 * 1 + 2 * 2 = 5
Example 2:
Input: 3
Output: False
题目大意：给一个数c，判断是否存在两个数字a b满足a*a+b*b=c，存在就返回true，否则返回false
分析：i和j分别从根号c开始到0，如果找到了i * i + j * j == c就return true，如果i * i + j * j < c就break第二层for循环，最后如果都没有找到就返回false～
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class Solution { public: bool judgeSquareSum(int c) { int t = sqrt(c); for (int i = t; i >= 0; i--) { for (int j = t; j >= 0; j--) { if (i * i + j * j == c) return true; if (i * i + j * j < c) break; } } return false; } }; |