1、題目

Given an array and a value, 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 in place with constant memory.

The order of elements can be changed. It doesn't matter what you leave beyond the new length.

Example:
Given input array?nums?=?[3,2,2,3],?val?=?3

Your function should return length = 2, with the first two elements of?nums?being 2.

?

2、代碼實現(xiàn)

java: public class Solution {public int removeElement(int[] nums, int val) {if (nums == null || nums.length == 0)return 0;int count = 0;for (int i = 0; i < nums.length; ++i) {if (nums[i] != val) {nums[count++] = nums[i];}}return count;} }
? C++: class Solution { public:int removeElement(vector<int>& nums, int val) {int count = 0;for (int i = 0; i < nums.size(); ++i) {if (nums[i] != val) {nums[count++] = nums[i];}}return count;} };
?

3、總結(jié)

其實非常簡單，把不相等的數(shù)據(jù)從頭到尾放進數(shù)組就行

總結(jié)

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