[LeetCode] 27. Remove Element 移除元素

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]);
}

這道題讓我們移除一個(gè)數(shù)組中和給定值相同的數(shù)字，并返回新的數(shù)組的長(zhǎng)度。是一道比較容易的題，只需要一個(gè)變量用來(lái)計(jì)數(shù)，然后遍歷原數(shù)組，如果當(dāng)前的值和給定值不同，就把當(dāng)前值覆蓋計(jì)數(shù)變量的位置，并將計(jì)數(shù)變量加1。代碼如下：

class Solution {
public:
    int removeElement(vector<int>& nums, int val) {
        int res = 0;
        for (int i = 0; i < nums.size(); ++i) {
            if (nums[i] != val) nums[res++] = nums[i];
        }
        return res;
    }
};

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