示例 1: code class Solution { public int numWaterBottle 阅读全文 » 2021215 LeetCode刷题 比特位计数(难度 :单词规律) 发表于 2021-12-15 20:41阅读:28评论:0推荐:0 摘要:题目: 给你一个整数 n ,对于 0 <= i <= n 中的每个 i ,计算其二进制表示中 1 的个数 ,返回一个长度为 n + 1...
class Solution { public: int assignBikes(vector<vector<int>>& workers, vector<vector<int>>& bikes) { int m = workers.size(), n = bikes.size(), i, j; vector<vector<int>> dis(m, vector<int>(n)); for(i = 0; i < m; ++i) for(j = 0; j < n; ++j) dis[i][j] = ...
1700-1799 1768. Merge Strings Alternately 交替合并字符串.md 1769. Minimum Number of Operations to Move All Balls to Each Box 移动所有球到每个盒子所需的最小操作数.md 200-299 200. Number of Islands 岛屿数量.md 201. Bitwise AND of Numbers Range 数字范围按位与.md 202. Happy Number 快乐...
继续化简,xi+1=xi- (xi2- n) / (2xi) = xi- xi/ 2 + n / (2xi) = xi/ 2 + n / 2xi= (xi+ n/xi) / 2。 有了迭代公式,程序就好写了。关于牛顿迭代法,可以参考wikipedia以及百度百科。 1intsqrt(intx) {2if(x ==0)return0;3doublelast =0;4doubleres =1;5while(res !=last)6{7...
我的解决方案: class Solution { public int maxProfit(int[] prices) { if(prices.length==0) return 0; //用后面的减去前面的,随时更新前面的值 //最后得到一个最大值 int front; front=prices[0]; int res=0; for(int i=1;i<prices.length;i++){ ...
class Solution { bool ans = false; public: bool canDistribute(vector<int>& nums, vector<int>& quantity) { unordered_map<int,int> m; for(auto n : nums) m[n]++; vector<int> val(m.size(), 0); int i = 0; for(auto it = m.begin(); it != m.end(); ++it) val[i++] =...
classSolution{public:vector<int>maximumBobPoints(int numArrows,vector<int>&aliceArrows){int maxscore=0,ans_state=0;for(int state=1;state<(1<<12);++state){int needarrows=0,score=0;for(int j=0;j<12;++j){if((state>>j)&1)// bob 要取得 j 的得分{needarrows+=aliceArrows[j]+1;/...
classSolution{int ans=0;public:intsubsetXORSum(vector<int>&nums){dfs(nums,0,0);returnans;}voiddfs(vector<int>&nums,int i,intXOR){if(i==nums.size()){ans+=XOR;return;}dfs(nums,i+1,XOR);dfs(nums,i+1,XOR^nums[i]);}};