leetcode--线性表--数组专题

最新推荐文章于 2022-07-30 00:19:02 发布

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本文深入探讨了在数组中寻找唯一出现一次的元素的算法，包括单次出现和三次出现的情况，并提供了高效的解决方案。同时，文章还介绍了糖果分配问题、加油站问题、矩阵置零问题等算法题目的解决思路。

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1. Single Number 1

题目描述：
given an array of integres, every elements appears twice except for one ,find that single one
Note: your algorithm should have a linear runtime complexity,could you implement it without using extra memory?
采用异或，不仅能处理两次的情况，只要出现偶数次，都可以清零

#include<iostream>
#include<stdio.h>
#include<vector>
using namespace std;
int singleNumber(vector<int>& nums)
{
    int x = 0;
    for (int i : nums)
    {
        x ^= i;
    }
    return x;

int main()
{
    int num;
    vector<int> v;
    while (cin >> num)
    {
        v.push_back(num);
        if (cin.get() == '\n')
        {
            cout << singleNumber(v) << endl;
            v.clear();
        }       

    }

}

2. single number 2

题目描述：
given an array of integres, every elements appears three time except for one ,find that single one
Note: your algorithm should have a linear runtime complexity,could you implement it without using extra memory?
count[i]表示在第i位出现1的次数，如果count[i]是3的整数倍，则忽略，否则取出来组成答案

int singleNumber(vector<int>& nums)
{
    const int w = sizeof(int) * 8;  //一个整数的bit位
    int count[w];  //在i位出现的1的次数
    fill_n(&count[0],w,0); 
    for (int i = 0; i < nums.size(); i++)
    {
        for (int j = 0; j < w; j++)
        {
            count[j] += (nums[i] >> j) & 1;
            count[j] %= 3;   
        }
    }
    int result = 0;
    for (int i = 0; i < w; i++)
    {
        result += (count[i] << i);
    }
    return result;
}

3. Candy

#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<vector>
using namespace std;
int candy(const vector<int>& ratings)
{
    vector<int> f(ratings.size());
    int sum = 0;
    for (int i = 0; i < ratings.size(); i++)
    {
        sum += solve(ratings, f, i);
    }
    return sum;
}
int solve(const vector<int>& ratings, vector<int>& f, int i)
{
    if (f[i] == 0)
    {
        f[i] = 1; //跟它的邻居作比较
        if (i > 0 && ratings[i] > ratings[i - 1])
            f[i] = max(f[i], solve(ratings, f, i - 1) + 1);
        if (i<ratings.size() - 1 && ratings[i] >ratings[i + 1])
            f[i] = max(f[i], solve(ratings, f, i + 1) + 1);
    }
    return f[i];
}

4. Gas station

int canCompleteCircuit(vector<int>& gas, vector<int> &cost)
{
    //设置两个变量
    int total = 0;
    int j = -1;
    for (int i = 0, sum = 0; i < gas.size(); i++)
    {
        sum += gas[i] - cost[i]; //判断当前指针的有效性
        total += gas[i] - cost[i]; //判断整个数组是否有解
        if (sum < 0)
        {
            j = i;
            sum = 0;
        }
    }
    return total >= 0 ? j + 1 : -1;
}

5. set matrix zeros

o(mn)—o(m+n)—常数空间

//如果某一行或者某一列为0.将整行或者整列都置0
//空间复杂度为o(1)
void setZeroes(vector<vector<int>>& matrix)
{
    const int m = matrix.size();
    const int n = matrix[0].size();
    bool row_zero = false;
    bool col_zero = false;//第一列是否存在0
    //第一行和第一列标记是否有0
    for (int i = 0; i < n; i++)
    {
        if (matrix[0][i] == 0)
        {
            row_zero = true;
            break;
        }
    }
    for (int i = 0; i < m; i++)
    {
        if (matrix[i][0] == 0)
        {
            col_zero = true;
            break;
        }
    }
    //复用第一行第一列
    for (int i = 1; i < m; i++)
    {
        for (int j = 1; j < n; j++)
        {
            if (matrix[i][j] == 0)
            {
                //借助第i行和第j列标记第一行和第一列
                matrix[0][j] = 0;
                matrix[i][0] = 0;
            }
        }
    }
    //利用第一行和第一列对第i行和第j列进行标记
    for (int i = 1; i < m; i++)
    {
        for (int j = 1; j < n; j++)
        {
            if (matrix[i][0] == 0 || matrix[0][j] == 0)
                matrix[i][j] = 0;
        }
    }
    if (row_zero)
        for (int i = 0; i < n; i++)
            matrix[0][i] = 0;
    if (col_zero)
        for (int i = 0; i < m; i++)
            matrix[i][0] = 0;

}