Leetcode27.Remove Element

27.Remove Element

題目

• 原題鏈接：https://leetcode.com/problems/remove-element/
• 題目內容：

     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]);
    }

解題思路

• 方案一
  最簡單的思路是遍歷數組nums，遍歷到與val相等的數組元素位置i，則數組從位置i到數組尾部元素均向前移動一位。時間複雜度是O(n²)。注意特殊情況，數組最後一個元素等於val時不再移動元素，而是刪除數組最後一個元素。
• 方案二
  通過雙指針實現，當nums[i]==val時 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;
   }
};