classSolution:deffindRepeatNumber(self, nums: List[int]) ->int: dic={}foriinnums:ifinotindic: dic[i]=1else:returnireturnFalse 思路:2. 数组哈希法,遍历数组,把序列[2,3,1,0,2,5,3]修改成一个下标和下标对应值是相同的数组[0,1,2,3,2,5,3] (nums[nums[i]] = nums[i]),寻找当前位(...
// Solution 1: Keep the statistics of number of nodes for every node in the tree. If you need statistics of a data structure which changes quite often, don't try to bind them together too tight. Use things like timestamps to check if the results in cache needs updating. It's a cac...
五、参考代码 class Solution { public int findRepeatNumber(int[] nums) { //设索引初始...
public int findRepeatNumber(int[] nums) { Set<Integer> dic = new HashSet<>(); for(int num : nums) { if(dic.contains(num)) return num; dic.add(num); } return -1; } } 作者:jyd 链接:https://leetcode.cn/problems/shu-zu-zhong-zhong-fu-de-shu-zi-lcof/solution/mian-shi-ti-03...
class Solution { public: int findRepeatNumber(vector<int>& nums) { int n = nums.size(); sort(nums.begin(), nums.end()); for(int i=0; i<n; ++i) { if(nums[i] == nums[i+1]) { return nums[i]; } } return -1; } }; 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11....
Could you do it in one-pass, using only O(1) extra memory and without modifying the value of the board? 【解答】要数有多少 battleship,并且要求使用 O(1) 的空间复杂度,还不能修改 board 上的数值。 一行一行遍历,每一行中从左往右遍历。对于每一个点,如果左侧和上方都不是 X,那就...
Each rectangle is defined by its bottom left corner and top right corner as shown in the figure. Assume that the total area is never beyond the maximum possible value of int. 【解答】这个题目主要需要注意的是几种矩形放置情况,考虑二者位置之间的关系。先计算两个矩形的面积之和,考虑到计算过程中...
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We can also project the arrays to a new array with length to be the largest element in the array. Then iterate over the array and get the longest consecutive sequence. If the largest number is very large, then the time complexity would be bad. ...