[LeetCode] 27. Remove Element

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]);
}

---

写了一个直接调earse和remove函数的，不贴出来了，太偷懒了,贴一个正常点的

把与val相等的元素移到vector末尾，j保存最后一个不是val的下标，用于跟和val相等的元素交换

int removeElement(vector<int>& nums, int val) {
    int i = 0, j = nums.size()-1;
    while (i <= j)
    {
        if (nums[j] == val)
        {
            --j;
        }
        else
        {
            swap (nums[i], nums[j]);
            ++i;
        }
    }

    return j + 1;
}

LeetCode上其他代码，更加简洁，思路也不一样

用新的值直接覆盖旧的值，等于val的丢掉

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