letecode [27] - Remove Element
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 ofnums
containing0
,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.
题目大意:
给定一个无序数组,删除与指定元素相同的所有元素,返回操作后的数组长度。
理 解 :
该题类似于上个删除重复元素的题目。方法也类似。
设置两个“指针” ,j 指针用来遍历比较判断是否满足条件, i 指针用来保存 新的数组元素(即未删除的元素)。
代 码 C++:
class Solution { public: int removeElement(vector<int>& nums, int val) { int n = nums.size(); if(n==0) return 0; int count=0,i,j; for(i=0,j=0;i<n && j<n;++i){ while(j<n && nums[j]==val) ++j; if(j>=n) break; nums[i] = nums[j++]; ++count; } return count; } };
运行结果:
执行用时 : 8 ms 内存消耗 : 8.7 MB