LeetCode 27. Remove Element
Description
Given an array and a value, remove all instances of that value in-place and return the new length.
Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory.
The order of elements can be changed. It doesn’t matter what you leave beyond the new length.
class Solution {
public int removeElement(int[] nums, int val) {
int i = 0;
for (int j = 0; j < nums.length; j++) {
if (nums[j] != val) {
nums[i++] = nums[j];
}
}
return i;
}
}
Complexity analysis
Time complexity : O(n). Assume the array has a total of n elements, both i and j traverse at most 2n steps.
Space complexity : O(1).
本文解析了LeetCode第27题Remove Element的解决方案,介绍了如何通过修改输入数组来移除特定元素,并保持O(1)的空间复杂度。文章详细阐述了双指针技巧的应用,同时提供了完整的代码实现及时间、空间复杂度分析。
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