算法设计与应用基础-第十三周&十四周

Range Sum Query - Immutable

Given an integer array nums, find the sum of the elements between indicesi andj (i ≤j), inclusive.

Example:

Given nums = [-2, 0, 3, -5, 2, -1]sumRange(0, 2) -> 1sumRange(2, 5) -> -1sumRange(0, 5) -> -3

class NumArray {private:    vector<int> sums;public:    NumArray(vector<int> nums) {        int sum=0;        for(int i=0;i<nums.size();i++)        {            sum += nums[i];            sums.push_back(sum);        }    }        int sumRange(int i, int j) {        return (sums[j]-sums[i-1]);    }};/** * Your NumArray object will be instantiated and called as such: * NumArray obj = new NumArray(nums); * int param_1 = obj.sumRange(i,j); */

Counting Bits

Given a non negative integer number num. For every numbersi in the range0 ≤ i ≤ num calculate the number of 1's in their binary representation and return them as an array.

Example:
For num = 5 you should return[0,1,1,2,1,2].

0000    0-------------   0001    1-------------   0010    1   0011    2-------------   0100    1   0101    2   0110    2   0111    3-------------   1000    1   1001    2   1010    2   1011    3   1100    2   1101    3   1110    3   1111    4

从1开始，遇到偶数时，其1的个数和该偶数除以2得到的数字的1的个数相同，遇到奇数时，其1的个数等于该奇数除以2得到的数字的1的个数再加1

class Solution {public:    vector<int> countBits(int num) {        vector<int> res;        res.push_back(0);        for(int i=1;i<=num;i++)        {            int count=res[i/2]+i%2;            res.push_back(count);        }        return res;}};

Continuous Subarray Sum

Given a list of non-negative numbers and a targetinteger k, write a function to check if the array has a continuous subarray of size at least 2 that sums up to the multiple ofk, that is, sums up to n*k where n is also aninteger.

Example 1:

Input: [23, 2, 4, 6, 7],  k=6Output: TrueExplanation: Because [2, 4] is a continuous subarray of size 2 and sums up to 6.

Example 2:

Input: [23, 2, 6, 4, 7],  k=6Output: TrueExplanation: Because [23, 2, 6, 4, 7] is an continuous subarray of size 5 and sums up to 42.

不断遍历获取不同的子数组之和并判断是否可以整除K，若整除即返回，若遍历完后都没有满足条件的就返回错误。要注意K=0时候的情况。

#include <vector>class Solution {public:    bool checkSubarraySum(vector<int>& nums, int k) {        vector<int> sums;        int sum=0;        for(int i=0;i<nums.size();i++)        {            sum += nums[i];            sums.push_back(sum);        }        for(int i=0;i<nums.size();i++)        {            for(int j=i+1;j<nums.size();j++)            {                if((k!=0&&(sums[j]-sums[i]+nums[i])%k==0)||(k==0&&(sums[j]-sums[i]+nums[i])==0))                    return true;            }        }        return false;    }};