Description Given an array of integers, remove the duplicate numbers in it. You should: Do it in place in the array. Move the unique numbers to the front of the array. Return the total number of the unique numbers. Example 1: Input:nums = [1,3,1,4,4,2]Output:[1,3,4,2,?,?]...
1. HashSet+swap publicclassSolution {/** @param nums: an array of integers * @return: the number of unique integers*/publicintdeduplication(int[] nums) {//write your code hereSet<Integer> set =newHashSet<Integer>();intcountDup = 0;for(inti = 0; i < nums.length - countDup; i++...
You must not modify the array (assume the array is read only). You must use only constant, O(1) extra space. Your runtime complexity should be less thanO(n2). 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...
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Note: You must not modify the array (assume the array is read only). You must use only constant, O(1) extra space. Your runtime complexity should be less than O(n2). There is only one duplicate number in the array, but it could be repeated more than once. ...
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- There is only one duplicate number in the array, but it could be repeated more than once. 代码如下: public int findDuplicate(int[] nums) { if (nums.length > 1){ int slow = nums[0]; int fast = nums[nums[0]]; //以下为求两个指针第一次相遇的点 while (slow != fast){ slow...
There is only one duplicate number in the array, but it could be repeated more than once. 解题思路 将数组的索引作为地址,存入数组的元素作为值,每个数组单元作为一个节点 由于不存在值为 0 的节点,因此构成的节点一定不会在首节点形成环 由于值中存在重复元素,因此构成的链表一定存在重复的节点指向同一个...
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