[学习报告]《LeetCode零基础指南》(第四讲) 数组

2021-11-25 22:31:20 124 阅读 AI 写同款 · GEO 优化AI 选题数据监测一键分发 ›

RK3576的ISP和VPSS图像处理能力实测：对比海思方案能做什么？本文深度评测了RK3576芯片的ISP和VPSS图像处理能力，并与海思方案进行对比。测试显示，RK3576在MIPI接口灵活性、多路视频支持、HDR动态范围和降噪效果等方面表现优异，特别适合智能视觉和安防监控应用。通过实测数据展示了其在低照度、高动态范围等复杂场景下的卓越性能。阅读详情

一.33. 搜索旋转排序数组
这三道数组题虽然会基本二分查找但是没有完全看明白题意，就先没管旋转了，就不用二分

class Solution {
public:
    int search(vector<int>& nums, int target) {
        int index = -1;
        for (int i = 0; i < nums.size(); i++) {
            if (nums[i] == target) {
                return i;
            }
        }
        return index;
    }
};

二.搜索旋转排序数组 II

class Solution {
public:
    bool search(vector<int>& nums, int target) {
        
        for (int i = 0; i < nums.size(); i++) {
            if (target == nums[i]) {
                return true;
            }
        }
        return false;
    }
};

三.寻找旋转排序数组中的最小值

class Solution {
public:
    int findMin(vector<int>& nums) {
        int min = 5000;
        for (int i = 0; i < nums.size(); i++) {
            if (nums[i] < min) {
                min = nums[i];
            }
        }
        return min;
    }

};
四.爬楼梯
递归超时,
今天学会了一种新方式动态的数组那样因为结果是只用前二个，前面的所有系列后面不用放在就浪费空间了

//第一种

class Solution {
public:
    int climbStairs(int n) {
        if (n <= 2) {return n;}
        int arr[1000] = {0};
        arr[1] = 1,arr[2] = 2;
        int i;
        for (i=3; i <= n; i++) {
            arr[i] = arr[i-1] + arr[i-2];
        }
        return arr[i-1];
        
    }

};
//第二种
class Solution {
public:
    int climbStairs(int n) {
        if (n <= 2) {return n;}
        int a = 1,b = 2;
        int i;
        for (i=3; i <= n; i++) {
           int c = a + b;
           a = b;
           b = c;
        }
        return b;
    }
};

五.斐波那契数

class Solution {
public:
    int fib(int n) {
        if(n < 2) return n;
        int dp[2];
        dp[0] = 0,dp[1] = 1;
        for (int i = 2; i <= n; i++) {
            int sum = dp[0] + dp[1];
            dp[0] = dp[1];
            dp[1] = sum;
        }
        return dp[1];
    }
};

class Solution {
public:
    int fib(int n) {
        if(n < 2) return n;
        return fib(n-1) + fib(n-2);
    }
};

class Solution {
public:
    int fib(int n) {
        if(n < 2) return n;
        int arr[31] = {0};
        arr[1] = 1;
        int i;
        for ( i = 2; i<=n; ++i) {
            arr[i] = arr[i-1] + arr[i-2];
        }
        return arr[i-1];
    }
};

六.第 N 个泰波那契数

class Solution {
public:
    int tribonacci(int n) {
        if (n == 0) {
            return 0;
        } else if (n == 1 || n == 2){
            return 1;
        }
        int dp[3];

        dp[0] = 0,dp[1] = 1,dp[2] = 1;
        for (int i = 3; i <= n; i++) {
            int sum = dp[0] + dp[1] + dp[2];
            dp[0] = dp[1];
            dp[1] = dp[2];
            dp[2] = sum;
        }
        return dp[2];
    }
};

七.差的绝对值为 K 的数对数目

class Solution {
public:
    int countKDifference(vector<int>& nums, int k) {
        int cnt = 0;
        for (int i = 0; i < nums.size(); i++) {
            for (int j = i+1; j < nums.size(); j++) {
                if (abs(nums[i] - nums[j]) == k) {
                    ++cnt;
                }
            }
        }
        return cnt;
    }
};

八.LCP 01. 猜数字

class Solution {
public:
    int game(vector<int>& guess, vector<int>& answer) {
        int cnt = 0;
        int i;
        for (i = 0; i<3; i++) {
            if(guess[i] == answer[i]) {++cnt;}
        }
        return cnt;
    }
};
//三目我觉得好看点
class Solution {
public:
    int game(vector<int>& guess, vector<int>& answer) {
        int cnt = 0;
        int i;
        for (i = 0; i<3; i++) {
            cnt += (guess[i] == answer[i]) ? 1 : 0;
        }
        return cnt;
    }
};

九.拿硬币

一开始用判断奇数偶数后面发现int 偶数+1/2次数还是一样

class Solution {
public:
    int minCount(vector<int>& coins) {
        int cnt = 0;
        int i;
        for (i = 0; i<coins.size(); i++) {
            cnt += (coins[i] + 1) / 2;
        }
        return cnt;
    }
};