【leetcode】27. Remove Element
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一、题目描述
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.
Hint:
- Try two pointers.
题目解读:给一个序列和一个值,去掉序列中的所有等于这个值的元素,返回新序列的长度
思路:设置两个指针,一个从头往后遍历,一个从后往前遍历。从头开始的指针只要遇到要去掉的元素就和尾的指针指向的元素交换位置。这样当start与end相遇时,前面的就是新序列,后面的就是要去掉的元素。返回start即可。
c++代码(4ms,13.35%)
class Solution {public: int removeElement(vector<int>& nums, int val) { if(nums.size() == 0) return 0; int start = 0; int end = nums.size()-1; while(start <= end){ if(nums[start] == val){ swap(nums[start], nums[end--]); }else{ start++ ; } } return start; }};
其他代码:
int removeElement(vector<int>& nums, int val) { int cnt = 0; for(int i = 0 ; i < nums.size() ; ++i) { if(nums[i] == val) cnt++; else nums[i-cnt] = nums[i]; } return nums.size()-cnt;}
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