[LeetCode]27. Remove Element

Problem

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.

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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:

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// 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]);
}

给定一个数组nums以及一个整数val, 就地删除等于value的元素,然后返回新的数组长度.

不要分配独立的新空间, 要求空间复杂度为O(1).数组中元素的位置可以改变. 数组长度之外的元素是什么无关紧要.

Solution

Analysis

这道题和删除数组中的重复元素解法类似. 使用双指针法. 这里是数组元素和value进行比较, 不相等,添加到新数组中.

Code

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class Solution {
public:
int removeElement(vector<int>& nums, int val) {
int idx = -1;

for (int i=0; i< nums.size(); i++){
if (nums[i] != val)
nums[++idx] = nums[i];
}

return idx+1;
}
};
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