leetcode027-python 移除元素

最新推荐文章于 2020-12-07 18:02:23 发布

MonkyK

最新推荐文章于 2020-12-07 18:02:23 发布

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分类专栏： leetcode 文章标签： leetcode-简单

本文链接：https://blog.youkuaiyun.com/chkay399/article/details/80795403

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本文介绍了一种在不使用额外空间的情况下从数组中移除特定值的方法，并保持了原数组的有效元素。通过两个指针i和j进行扫描，将等于目标值的元素移到数组后部，不等于目标值的元素则保留在前部。该方法符合O(1)额外空间的要求。

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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]);
}

解题要点：

1.把等于val的值放到后面，后面不等于val的值放到前面val处，第一种做法有些复杂

class Solution:
    def removeElement(self, nums, val):
        """
        :type nums: List[int]
        :type val: int
        :rtype: int
        """
        n = len(nums)
        if n == 0:
            return n

        i, j = 0 , n-1
        while i <= j:
            while nums[i] != val:
                i += 1
                if i > j:
                    break
            while nums[j] == val:
                j -= 1
                if i > j:
                    break
            if i > j:
                    break
            nums[i] = nums[j]

            i += 1
            j -= 1
        return i