LeetCode27 Remove Element

问题描述：
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]);
}
题源：here；完整代码：here
思路：
两种方案：1 依次替换；2 只替换和val相等的数。两种方案复杂度相同，但是因为第二种方案会减少赋值次数，所以效率高一些。
方案1
同上一题处理类似，此题，我们将那些不相同的数依次压入数组前面，这样的不会改变输入数据的顺序。
代码

    int removeElement_1(vector<int>& nums, int val) {
        if (!nums.size()) return 0;
        int len = 0;
        for (int i = 0; i < nums.size(); i++){
            if (nums[i] == val) continue;
            else nums[len++] = nums[i];
        }

        return len;
    }

方案2
因为题目不要求保持输入数据的顺序，所以我们将与val相等的数和最后的数调换位置，然后重新判断（考虑最后的数和val相同的情况）。
代码

    int removeElement_2(vector<int>& nums, int val){
        if (!nums.size()) return 0;
        int start = 0, end = nums.size()-1;
        while (start <= end){
            if (nums[start] == val) nums[start] = nums[end--];
            else start++;
        }

        return end + 1;
    }