标签:algorithm leetcode python c++
Given an array and a value, remove all instances of that value in place and return the new length.
The order of elements can be changed. It doesn‘t matter what you leave beyond the new length.
class Solution { public: int removeElement(vector<int>& nums, int val) { int n = nums.size(); int i = 0; while( i < n ) { if( nums[i] == val ) { swap(nums[i], nums[n-1]); n--; } else { i++; } } return n; } };
int removeElement(vector<int>& nums, int val) { vector<int>::iterator itr = nums.begin(); while (itr != nums.end()) { if (*itr == val) itr = nums.erase(itr); else ++itr; } return nums.size(); }
int removeElement(int A[], int n, int elem) { int begin=0; for(int i=0;i<n;i++) if(A[i]!=elem) A[begin++]=A[i]; return begin; }
class Solution: # @param A a list of integers # @param elem an integer, value need to be removed # @return an integer def removeElement(self, A, elem): i = 0 for j in range(len(A)): if A[j] != elem: A[i] = A[j] i += 1 return i
def removeElement(self, nums, val): nums[:] = [x for x in nums if x!=val] return len(nums)
标签:algorithm leetcode python c++
原文地址:http://blog.csdn.net/wangyaninglm/article/details/46622949