Leetcode27.Remove Element

27.Remove Element

题目

• 原题链接：https://leetcode.com/problems/remove-element/
• 题目内容：

     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 of nums containing 0, 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.
    Clarification:
    
    Confused why the returned value is an integer but your answer is an array?
    
    Note that the input array is passed in by reference, which means modification to the input array will be known to the caller as well.
    
    Internally you can think of this:
    
    // nums is passed in by reference. (i.e., without making a copy)
    int len = removeElement(nums, val);
    
    // any modification to nums in your function would be known by the caller.
    // using the length returned by your function, it prints the first len elements.
    for (int i = 0; i < len; i++) {
        print(nums[i]);
    }

解题思路

• 方案一
  最简单的思路是遍历数组nums，遍历到与val相等的数组元素位置i，则数组从位置i到数组尾部元素均向前移动一位。时间复杂度是O(n²)。注意特殊情况，数组最后一个元素等于val时不再移动元素，而是删除数组最后一个元素。
• 方案二
  通过双指针实现，当nums[i]==val时 nums[j++] = nums[i]。

class Solution {
public:
   int removeElement(vector<int>& nums, int val) {
       int length = nums.size();
       int i, j=0, count=0;
       for(i=0;i<length;++i){
           if(nums[i]==val)
               ++count;
           else
               nums[j++] = nums[i];
       }
       return length - count;
   }
};