代码随想录day2 数组总结

1.二分查找

2.快慢 双指针

代码随想录day1-CSDN博客

3.滑动窗口

滑动窗口就是有一个起始位置，一个终止位置，通过调节起始位置和终止位置得到我们想要的结果。

外面一层for循环    用来更新终止位置   不满足条件 终止位置右移

里面一层while循环  用来更新起始位置   满足条件 起始位置右移

209. 长度最小的子数组 - 力扣（LeetCode）

int minSubArrayLen(int target, vector<int>& nums) {
        int i = 0;
        int reslut = 100001;
        int subL = 0;
        int sum = 0;

        for(int j = 0; j < nums.size(); j++){
            sum += nums[j];
            while(sum >= target){
                subL = j - i + 1;
                reslut = min(reslut, subL);
                sum -= nums[i];
                i++;
            }
        }
        if(reslut == 100001)return 0;
        else return reslut;
    }

904. 水果成篮 - 力扣（LeetCode）

int totalFruit(vector<int>& fruits) {
        int i = 0;
        int result = 0;

        unordered_map<int, int> kind; //记录水果种类数

        for(int j = 0; j < fruits.size(); j++){
            kind[fruits[j]]++;

            while(kind.size() > 2){
                kind[fruits[i]]--;
                if(kind[fruits[i]] == 0){
                    kind.erase(fruits[i]); //如果某种水果数量等于0，从哈希表中移除
                }
                i++;
            }

            result = max(result, j - i + 1);
        }

        return result;
    }

76. 最小覆盖子串 - 力扣（LeetCode）

string minWindow(string s, string t) {
        if(s.empty() || t.empty() || s.size() < t.size())return "";
        
        int result = INT_MAX;
        int i = 0;

        unordered_map<char, int>t_count;
        unordered_map<char, int> window_count;

        for(int j = 0; j < t.size(); j++){
            t_count[t[j]]++;
            
        }

        int num = 0;
        int left = 0;
        for(int j = 0; j < s.size(); j++){
            window_count[s[j]]++;
            if(window_count[s[j]] == t_count[s[j]]){
                num++;
            }

            while(num == t_count.size()){
                int temp = j - i + 1;
                if(temp < result){
                    result = temp;
                    left = i;
                }
                if(t_count.find(s[i]) != t_count.end()){
                    if(window_count[s[i]] == t_count[s[i]]){
                        num--;
                    }//本来有可能大于或等于
                    window_count[s[i]]--;
                }
                i++;
            }
        }

        return result == INT_MAX ? "":s.substr(left, result);
    }

4.螺旋矩阵模拟

左闭右开，循环模拟

59. 螺旋矩阵 II - 力扣（LeetCode）

vector<vector<int>> generateMatrix(int n) {
        int starx = 0, stary = 0;
        int offset = 1;
        vector<vector<int>> re(n, vector<int>(n,0));
        int count = 1;
        int i, j;

        while(offset <= n/2){
            i = starx;
            j = stary;
            for(; j < n - offset; j++){
                re[i][j] = count++;
            }
            for(; i < n - offset; i++){
                re[i][j] = count++;
            }
            for(; j > stary; j--){
                re[i][j] = count++;
            }
            for(; i > starx; i--){
                re[i][j] = count++;
            }
            offset++;
            starx++;
            stary++;
        }
        if(n % 2 != 0)re[n/2][n/2] = count;

        return re;
    }

 54. 螺旋矩阵 - 力扣（LeetCode）

vector<int> spiralOrder(vector<vector<int>>& matrix) {
        if(matrix.size() == 0 || matrix[0].size() == 0)return {};

        vector<int> re; 
        int starx = 0, stary = 0;
        int w = matrix[0].size();
        int h = matrix.size();
        int i, j;

        int offset = 1;

        while(offset <= min(w,h)/2){
            i = starx;
            j = stary;
            for(; j < w - offset; j++){
                re.push_back(matrix[i][j]);
            }
            for(;i < h - offset; i++){
                re.push_back(matrix[i][j]);
            }
            for(; j > stary; j--){
                re.push_back(matrix[i][j]);
            }
            for(; i > starx; i--){
                re.push_back(matrix[i][j]);
            }
            offset++;
            starx++;
            stary++;
        }
        if(h % 2 != 0 && w >= h){
            int offset = h/2;
            for(int j = offset; j < w - offset; j++){
                re.push_back(matrix[offset][j]);
            }
        }else if(w % 2 != 0 && w < h){
            int offset = w/2;
            for(int i = offset; i < h - offset; i++){
                re.push_back(matrix[i][offset]);
            }
        }

        return re;
    }

5.区间和

把前缀和保存起来后面直接用

58. 区间和（第九期模拟笔试）

#include<iostream>
#include<vector>
using namespace std;

int main(){
    int n, a, b;
    cin >> n;
    int re = 0;

    vector<int> vec(n);
    vector<int> sum(n);   // 前缀和

    for(int i = 0; i < n; i++){
        int t;
        cin >> t;

        vec[i] = t;
        if(i == 0)sum[i] = t;
        else sum[i] = t + sum[i-1];
    }

    while(cin >> a >> b){
        if(a == 0)re = sum[b];
        else re = sum[b] - sum[a-1];
        cout << re << endl;
    }

}

44. 开发商购买土地（第五期模拟笔试）

#include<iostream>
#include<vector>
#include <climits>
using namespace std;

int main(){
    int n, m;
    cin >> n >> m;
    int sum = 0;

    vector<vector<int>> vec(n, vector<int>(m, 0));
    vector<int> w(n, 0);  //横向
    vector<int> h(m, 0);  //纵向

    for(int i = 0; i < n; i++){
        for(int j = 0; j < m; j++){
            int t = 0;
            cin >> vec[i][j];
            sum += vec[i][j];
            w[i] += vec[i][j];
        }
    }

    for(int j = 0; j < m; j++){
        for(int i = 0; i < n; i++){
            h[j] += vec[i][j];
        }
    }

    int result = INT_MAX;
    int t = 0;
    
    for(int i  = 0; i < n; i++){
        t += w[i];
        result = min(result, abs(sum - t - t));
    }

    t = 0;
    for(int j = 0; j < m; j++){
        t += h[j];
        result = min(result, abs(sum - t - t));
    }

    cout << result << endl;
}