3分鐘作5道leetcode題上手leetcode刷題之路

時間 2019-11-05

標籤 3分 leetcode 上手之路简体版

原文原文鏈接

官網地址：leetcode-cn.comphp

在官網註冊後，點擊題庫頁面就能夠看到全部能夠刷的題目了 html

能夠按照難度進行篩選、建議從簡單難度的題上手

兩數之和: leetcode-cn.com/problems/tw…

問題描述 java

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        vector<int> res;
        bool resolve = false;
        for (int i = 0; i < nums.size() - 1; i ++) {
            for (int j = i + 1; j < nums.size(); j ++){
                if (nums[i] + nums[j] == target) {
                    res.push_back(i);
                    res.push_back(j);
                    resolve = true;
                    break;
                }
            }
            if (resolve) {
                break;
            }
        }
        return res;
    }
};
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而後提交就能夠看到統計信息了 c++

leetcode題的答案不止一種，能夠根據cpu和內存使用的反饋進行調優。編程

2的冪: leetcode-cn.com/problems/po…

class Solution {
public:
    bool isPowerOfTwo(long n) {
	    return n != 0 && (n & (n -1)) == 0;
    }
};
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能夠在題解界面查看其它同窗的解題思路 api

4的冪: leetcode-cn.com/problems/po…

class Solution {
public:
    bool isPowerOfFour(long n) {
	    return n != 0 && (n & 0xAAAAAAAA) == 0 && (n & (n -1)) == 0;
    }
};
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斐波拉契數: leetcode-cn.com/problems/fi…

class Solution {
public:
    int fib(int n) {
        if (n == 0) {
            return 0;
        }
        if (n <= 2) {
            return 1;
        }
        int a = 1, b = 1, i = 3, res = 0;
        while (i <= n) {
            res = a + b;

            a = b;
            b = res;
            i ++;
        }
        return res;
    }
};
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移除元素: leetcode-cn.com/problems/re…

class Solution {
public:
    int removeElement(vector<int>& nums, int val) {
        int find = 0;
        
        int j = nums.size();
        for (int i = 0; i < j; i ++) {
            cout << "i: " << i << " num:" << nums[i] << endl;
            if (nums[i] == val) {
                find ++;
                swap(nums[i], nums[j - 1]);
                j --;
                i --;
            }
        }
        
        return nums.size() - find;
    }
};
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