代码随想录笔记--数组篇

1--二分查找法

二分查找法用于有序数组的元素查找，一般可以分为左闭右闭 写法、左闭右开 写法、左开右闭写法，其中左闭右闭写法最常用；

#include <iostream>
#include <vector>

class Solution {
public:
    // 左闭右闭写法
    int search1(std::vector<int>& arr, int target){
        int left = 0, right = arr.size() - 1;
        while(left <= right){ // 因为是左闭右闭写法，当left == right时[left, right]也是合法的，因此left <= right
            int mid = left + (right - left) / 2;
            if(arr[mid] > target){
                right = mid - 1;
            }
            else if(arr[mid] < target){
                left = mid + 1;
            }
            else return arr[mid];
        }
        return -1; // 表示未找到
    }

    // 左闭右开写法
    int search2(std::vector<int>& arr, int target){
        int left = 0, right = arr.size();
        while(left < right){ // 因为是左闭右开写法，当left == right时[left, right)是非法的，因此left < right
            int mid = left + (right - left) / 2;
            if(arr[mid] > target){
                right = mid; // 右开
            }
            else if(arr[mid] < target){
                left = mid + 1; // 左闭
            }
            else return arr[mid];
        }
        return -1; // 表示未找到
    }

    // 左开右闭写法
    int search3(std::vector<int>& arr, int target){
        int left = -1, right = arr.size() - 1;
        while(left < right){ // 因为是左闭右开写法，当left == right时(left, right]是非法的，因此left < right
            int mid = left + (right - left) / 2;
            if(arr[mid] > target){
                right = mid - 1; // 右闭
            }
            else if(arr[mid] < target){
                left = mid; // 左开
            }
            else return arr[mid];
        }
        return -1; // 表示未找到
    }
};

int main(int argc, char *argv[]){
    std::vector<int> test = {1, 2, 3, 4, 5, 6, 7, 8, 9};
    int target = 2;
    Solution S1;
    int res1 = S1.search1(test, target);
    int res2 = S1.search2(test, target);
    int res3 = S1.search3(test, target);
    std::cout << res1 << ", " << res2 << ", " << res3 << std::endl;
    return 0;
}

2--二分查找法进阶

2-1--寻找第一个等于目标值的位置

等于号 = 放在哪个条件判断的依据：当 arrmid == target 时，应该往哪一边继续二分寻找；

#include <iostream>
#include <vector>

class Solution {
public:
    // 左闭右闭写法
    int search1(std::vector<int>& arr, int target){
        int left = 0, right = arr.size() - 1;
        int res = -1;
        while(left <= right){
            int mid = left + (right - left) / 2;
            if(arr[mid] >= target){
                res = mid; // 记录可能的位置
                right = mid - 1;
            }
            else if(arr[mid] < target){
                left = mid + 1;
            }
        }
        if(res == -1 || arr[res] != target) return -1; //未找到target
        return res;
    }

};

int main(int argc, char *argv[]){
    std::vector<int> test = {1, 2, 2, 2, 3, 3, 4};
    int target = 2;
    Solution S1;
    int res1 = S1.search1(test, target);
    std::cout << res1 << std::endl;
    return 0;
}

2-2--寻找最后一个等于目标值的位置

等于号 = 放在哪个条件判断的依据：当 arrmid == target 时，应该往哪一边继续二分寻找；

#include <iostream>
#include <vector>

class Solution {
public:
    // 左闭右闭写法
    int search1(std::vector<int>& arr, int target){
        int left = 0, right = arr.size() - 1;
        int res = -1;
        while(left <= right){
            int mid = left + (right - left) / 2;
            if(arr[mid] > target){
                right = mid - 1;
            }
            else if(arr[mid] <= target){
                res = mid; // 记录可能的位置
                left = mid + 1;
            }
        }
        if(res == -1 || arr[res] != target) return -1; //未找到target
        return res;
    }

};

int main(int argc, char *argv[]){
    std::vector<int> test = {1, 2, 2, 2, 3, 3, 4};
    int target = 2;
    Solution S1;
    int res1 = S1.search1(test, target);
    std::cout << res1 << std::endl;
    return 0;
}