leetcode解题: Remove Element (27)

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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.
Did you use the property of “the order of elements can be changed”?
What happens when the elements to remove are rare?

解法1：Two pointers

用two pointers，back记录需要remove的起始的位置。front不停的往后扫描，当扫到不要remove的时候则前进，如果扫到需要remove的就和back指向的数值交换，back向前-1
C++

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class Solution {
public:
    int removeElement(vector<int>& nums, int val) {
        if (nums.empty()) {
            return 0;
        }
        int front = 0, back = nums.size() - 1;
        while (front <= back) {
            if (nums[front] != val) {
                ++front;
            } else {
                int temp = nums[back];
                nums[back] = nums[front];
                nums[front] = temp;
                --back;
            }
        }
        return front;
    }
};

Java

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