Two-pointer approach: We use two pointers (left and right) that start at the beginning and end of the array, respectively. These pointers move towards the center. At each step, we calculate whether the left or right side has a lower height. We only calculate trapped water on the side th...
publicclassSolution {publicintthreeSumClosest(int[] num,inttarget) {intresult = num[0] + num[1] + num[num.length - 1];//初始化Arrays.sort(num);//选择一个数for(inti = 0; i < num.length - 2; i++) {intstart = i + 1, end = num.length - 1;//双指针指向最小数和最大数wh...
1.两个指针i,j,当i<j时,不断两头往里面扫描 2.每扫描一个位置计算当前的面积,并刷新最大的面积 3.ai和aj的小者继续往里面递进 时间复杂度O(n) intmaxArea(int* height,intheightSize) {if(NULL==height)return0;inti=0,j=heightSize-1;intarea,t_max=0;while(i<j){ area=(j-i)*(height[i]...
Assume that you are given a task to find a pair in an array that sums to S. Whenever I encounter problems around this line, the first approach that I can think of is theTwo-pointer approach. The reason behind that is it's super easy to understand and implement. How sorted array work?
LeetCode Two Pointers related problems before I wrote this article, I just think that two pointer technique is nothing like iteration, but in an efficient way. it’s like brute force with memo. sometimes, I mean, many times, I just implement those things straightly, and after that I found...
Leetcode: Intersection of Two Arrays II,用HashMapFollowUp1:用twopointer解,可以省存成HashMapFollowUp2:用在长的数组里面binarySearch可解FollowUp3:Whatifelementsofnums2arestoredondisk,andt
By using this strategy, we can efficiently eliminate duplicates from the sorted array without requiring any additional memory. The conditionnums[i] <= nums[i+1], inherent in the sorted nature of the array, facilitates the effectiveness of the two-pointer approach, leading to a more optimi...
if (currentNodeList1 != null) { sum += currentNodeList1.val; currentNodeList1 = currentNodeList1.next; } // current pointer of list 2 is not null, update the sum, and point current pointer to next node. if (currentNodeList2 != null) { sum += currentNodeList2.val; currentNodeList2...
if temp < minAbs: minAbs = temp # Adjust the pointer for next trying if back == front: break elif abs(A[back+1] + A[front]) <= temp: back += 1 elif abs(A[back] + A[front-1]) <= temp: front -= 1 else: back += 1; front -= 1 return minAbsCodility...
Solution: 思路: Solution1. 2sum on leftPath + rightPath with "two pointers" approach Solution2. in order 遍历 + 2sum (O(n) / O(n)) Time Complexity: O(N) Space Complexity: O(N) Avg Space Complexity: O(logH) Solution1 Code: ...