LeetCode 0027 - Remove Element
# Hints
- 水题
# 题面
Difficulty | Time Complexity Limit | Extra-Memory Complexity Limit |
---|---|---|
Easy |
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 extra memory.
The order of elements can be changed. It doesn't matter what you leave beyond the new length.
Example 1:
Given nums = [3,2,2,3], val = 3,
Your function should return length = 2, with the first two elements of nums being 2.
It doesn't matter what you leave beyond the returned length.
Example 2:
Given nums = [0,1,2,2,3,0,4,2], val = 2,
Your function should return length = 5, with the first five elements of nums containing 0, 1, 3, 0, and 4.
Note that the order of those five elements can be arbitrary.
It doesn't matter what values are set beyond the returned length.
Clarification:
Confused why the returned value is an integer but your answer is an array?
Note that the input array is passed in by reference, which means modification to the input array will be known to the caller as well.
Internally you can think of this:
// nums is passed in by reference. (i.e., without making a copy)
int len = removeDuplicates(nums);
// any modification to nums in your function would be known by the caller.
// using the length returned by your function, it prints the first len elements.
for (int i = 0; i < len; i++) {
print(nums[i]);
}
# 题意
给定一个数组 ,给定一个整数 ,要求将其中所有值为 的元素全部去除,通过引用修改 中的内容,并返回新的 的长度。新的 中的元素可按任意顺序摆放。
# 题解
水题,按题意模拟即可。
# AC代码
class Solution {
public:
int removeElement(vector<int> & nums, int val) {
int n = 0;
for (int i = 0; i < nums.size(); i ++) {
if (nums[i] != val) {
nums[n++] = nums[i];
}
}
return n;
}
};
- 01
- Reading Papers - Kernel Concurrency06-01
- 02
- Linux Kernel - Source Code Overview05-01
- 03
- Linux Kernel - Per-CPU Storage05-01