Given an array nums and a value val, remove all instances of that value in-place and return the new length.html
Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory.java
The order of elements can be changed. It doesn't matter what you leave beyond the new length.git
Example 1:github
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:數組
Given nums = [0,1,2,2,3,0,4,2], val = 2, Your function should return length = , with the first five elements of containing , , , , 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.5nums0130
Clarification:post
Confused why the returned value is an integer but your answer is an array?this
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.url
Internally you can think of this:spa
// 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]); }
這道題讓咱們移除一個數組中和給定值相同的數字,並返回新的數組的長度。是一道比較容易的題,只須要一個變量用來計數,而後遍歷原數組,若是當前的值和給定值不一樣,就把當前值覆蓋計數變量的位置,並將計數變量加1。代碼以下:code
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; } };
Github 同步地址:
https://github.com/grandyang/leetcode/issues/27
相似題目:
Remove Duplicates from Sorted Array
參考資料:
https://leetcode.com/problems/remove-element/
https://leetcode.com/problems/remove-element/discuss/12286/Accepted-java-solution
https://leetcode.com/problems/remove-element/discuss/12289/My-solution-for-your-reference.