第一题：

973. K Closest Points to Origin

 We have a list of  points on the plane.  Find the  K closest points to the origin  (0, 0).

(Here, the distance between two points on a plane is the Euclidean distance.)

You may return the answer in any order.  The answer is guaranteed to be unique (except for the order that it is in.)

 

Example 1:

Input: points = [[1,3],[-2,2]], K = 1 Output: [[-2,2]] Explanation: The distance between (1, 3) and the origin is sqrt(10). The distance between (-2, 2) and the origin is sqrt(8). Since sqrt(8) < sqrt(10), (-2, 2) is closer to the origin. We only want the closest K = 1 points from the origin, so the answer is just [[-2,2]].

Example 2:

Input: points = [[3,3],[5,-1],[-2,4]], K = 2 Output: [[3,3],[-2,4]] (The answer [[-2,4],[3,3]] would also be accepted.)

 

Note:

  1. 1 <= K <= points.length <= 10000
  2. -10000 < points[i][0] < 10000
  3. -10000 < points[i][1] < 10000

题目大意：给你一些点，让你找离远点最近的K个点。主要考的是二维数组排序。

class Solution {
public:
    vector<vector<int> > kClosest(vector<vector<int> >& points, int K) {
        vector<vector<int> > ans;
        int len = points.size();
        for(int i=0; i<len; i++) {
            int x = points[i][0];
            int y = points[i][1];
            points[i].push_back(x*x+y*y);
        }
        sort(points.begin(), points.end(), [](const vector<int> &a, const vector<int> &b) { return a[2] < b[2]; });
        for(int i=0; i<K; i++) {
            vector<int> res;
            res.push_back(points[i][0]);
            res.push_back(points[i][1]);
            ans.push_back(res);
        }
        return ans;
    }
};
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第二题：

976. Largest Perimeter Triangle

Given an array  A of positive lengths, return the largest perimeter of a triangle with non-zero area, formed from 3 of these lengths.

If it is impossible to form any triangle of non-zero area, return 0.

 

Example 1:

Input: [2,1,2]
Output: 5

Example 2:

Input: [1,2,1]
Output: 0

Example 3:

Input: [3,2,3,4]
Output: 10

Example 4:

Input: [3,6,2,3]
Output: 8

 

Note:

  1. 3 <= A.length <= 10000
  2. 1 <= A[i] <= 10^6

题目大意：从数组中，找出三个点组成一个周长最大的三角形，然后输出周长。

数据比较小，直接暴力的。

class Solution {
public:
    bool ok(int a, int b, int c) {
        return a-b<c;
    }

    int largestPerimeter(vector<int>& A) {
        int len = A.size();
        sort(A.begin(), A.end());
        for(int i=len-1; i>=0; i--) {
            for(int j=i-1; j>=0; j--) {
                if( ok(A[i], A[j], A[j-1])) {
                    return A[i]+A[j]+A[j-1];
                }
            }
        }
        return 0;
    }
};
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第三题：

974. Subarray Sums Divisible by K

Given an array  A of integers, return the number of (contiguous, non-empty) subarrays that have a sum divisible by  K.

 

Example 1:

Input: A = [4,5,0,-2,-3,1], K = 5 Output: 7 Explanation: There are 7 subarrays with a sum divisible by K = 5: [4, 5, 0, -2, -3, 1], [5], [5, 0], [5, 0, -2, -3], [0], [0, -2, -3], [-2, -3]

 

Note:

  1. 1 <= A.length <= 30000
  2. -10000 <= A[i] <= 10000
  3. 2 <= K <= 10000

题目大意：统计有多少条满足条件的子序列。条件1：连续，必须是连续子序列，条件2：序列和可以整除K。

思路：求出前缀和，想要找到连续子序列之和是可以整除K的话，那么前缀和只差模K也就是为0，也就是说统计模K相等的前缀和之差的个数，求一个等差数列求和。

做题的时候开始考虑错了，后面想到了解决方法但是没时间了，对照rank写出了一个我自己都认为是错的程序，提交AC。想不清楚，等过几天想清楚了，在仔细写下吧。

class Solution {
public:  
    int subarraysDivByK(vector<int>& A, int K) {
        int a[K];
        memset(a, 0, sizeof(a));
        int len = A.size(); a[0] = 1;
        int res = 0;
        for(int i=0; i<len; i++) {
            A[i] += A[i-1];
            int cnt = (A[i]%K+K)%K;
            a[cnt] ++;
        }
        for(int i=0; i<K; i++) {
            res += (a[i]-1)*a[i]/2;
        }
        return res;
    }
};
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