举例: nums = [4,5,6,7,0,1,2], target = 0 step 1: l, r = 0, 6, mid = 3, 此时 nums[3] 先跟 nums[l] 比较 如果 nums[3] > nums[...leetCode(search-in-rotated-sorted-array)-在一个旋转排序数组中查找目标值 题目:给定一个从中间某一位置翻转的排序数组,在该数组中查找目标值...
数之和 三数之和问题--leetcode 和两数之和问题类似,能想到的解法有两种: 解法一:暴力,判断所有的三数组合之和满不满足条件,枚举的复杂度为:空间:O(l), 时间:O(N^3) 解法二...1.0 寻找二数之和 两数之和问题--leetcode 即在数组nums[]中寻找和为target的元素下标。 题本身的解法很经典: 解法一:...
LeetCode-114. Flatten Binary Tree to Linked List Given a binary tree, flatten it to a linked list in-place. For example, Given 1 / \ 2 5 / \ \ 3 4 6 The flattened tree should look like: 1 \ 2 \ 3 \ 4 \ 5 \ 6 Hints: If you notice carefully in the flattened tree, each...
There is only one duplicate number in the array, but it could be repeated more than once 非常好的题目,开始是用二分做的,比如取数组为{1,2,3,3,4,5},mid应该是(5+1)/2 = 3,那么,如果小于等于mid的数的个数如果超过了3,那么重复的数字一定出现在l,mid之间,否则出现在mid + 1,r之间。以该...
https://leetcode.com/problems/linked-list-cycle-ii/solution/ Algorithm First off, we can easily show that the constraints of the problem imply that a cycle must exist. Because each number in nums is between 11 and nn, it will necessarily point to an index that exists. Therefore, the list...
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else left = mid + 1; mid = (left + right) / 2; }//find one position of the target digit if(nums[mid] != target) return new int[] {-1, -1}; int st = mid; int ed = mid; while(st >= 0 || ed < nums.length){ ...
Node tmp = leftIndex[mid]; if(tmp.value > rightIndex) { right = mid - 1; }else { left = mid + 1; } } if(leftIndex[right].value == rightIndex) { result[i] = leftIndex[right].index; }else if(right<intervals.length-1) { ...
1、如果num[mid] > num[mid + 1],说明在begin和mid 之间,至少存在一个peak element,因此只考察数组前半段就可以了; 2、如果num[mid] < num[mid + 1], 说明在mid + 1和end之间,至少存在一个peak element,因此只考察数组后半段就可以了;
解法2: 二分法。 每次比较的是mid位置和x的距离跟mid+k跟x的距离,以这两者的大小关系来确定二分法折半的方向,最后找到最近距离子数组的起始位置。 Java: 1 2 3 4 5 6 7 8 9 10 11 12 13 publicList<Integer> findClosestElements(int[] A,intk,intx) { ...