leetcode Day3----array.easy

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本文介绍了一种在原地删除数组中特定值的高效算法，通过修改输入数组而非使用额外空间，实现了O(1)的额外内存消耗。示例代码展示了如何在不改变剩余元素顺序的情况下，移除所有目标值并返回新数组长度。提供了C语言和Python3的实现方案，附带了运行时间和内存使用情况的性能分析。

Remove Element

Given an array nums and a value val, 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.

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 = removeElement(nums, val);

// 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]);
}

这个题和昨天的题类似，改改代码就好了。

C语言

int removeElement(int* nums, int numsSize, int val) {
    int m=0;
    for(int i=0;i<numsSize;i++){
        while(i<numsSize&&nums[i]==val)
            i++;
        if(i<numsSize){
            nums[m]=nums[i];
            m++;
        }
    }
    return m;
}

Success
Details
Runtime: 4 ms, faster than 87.56% of C online submissions for Remove Element.
Memory Usage: 6.9 MB, less than 100.00% of C online submissions for Remove Element.

python3

class Solution:
    def removeElement(self, nums: List[int], val: int) -> int:
        m=0
        for i in range(len(nums)):
            if(nums[i]!=val):
                nums[m]=nums[i]
                m+=1
        return m

Success
Details
Runtime: 36 ms, faster than 99.22% of Python3 online submissions for Remove Element.
Memory Usage: 13.1 MB, less than 5.09% of Python3 online submissions for Remove Element.
Next challenges: