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leetcode 27 Remove Element

时间:2015-06-24 16:20:38      阅读:228      评论:0      收藏:0      [点我收藏+]

Remove Element Total Accepted: 60351 Total Submissions: 187833 My Submissions
                     

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.

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c++ 解决方案:

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;
}


python解决方案:
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

史上最简洁的解决方案:
def removeElement(self, nums, val):
        nums[:] = [x for x in nums if x!=val]
        return len(nums)



leetcode 27 Remove Element

原文:http://blog.csdn.net/wangyaninglm/article/details/46622949

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