[LeetCode] 27. Remove Element 移除元素

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 = , with the first five elements of  containing , , , , 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.5nums0130

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]);
}

给定一个数组和一个值，移除全部和这个值相等的元素，返回新数组的长度。用in-place，O(1)空间。

解法：双指针，从第一个元素开始遍历，前面的指针i指向当前数，pos指针指向交换后的元素的下一位，若是和给定的数和当前数不等就和pos元素交换，last指针右移一位，继续比较。

Java:

public class Solution {
    public int removeElement(int[] nums, int val) {
        int position = 0;
        for (int i = 0; i < nums.length; i++) {
            if (nums[i] != val) {
                nums[position] = nums[i];
                position++;
            }
        }
        return position;
    }
}　

Python:

class Solution(object):
    def moveZero(self, nums, elem):
        i = 0
        for j in xrange(len(nums)):
            if nums[j] != elem:
                nums[i], nums[j] = nums[j], nums[i]
                i += 1
                
        return i　　

Python:

class Solution:
    def removeElement(self, A, elem):
        i, last = 0, len(A) - 1
        while i <= last:
            if A[i] == elem:
                A[i], A[last] = A[last], A[i]
                last -= 1
            else:
                i += 1
        return last + 1

C++:

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

　

变形：给一个字符数组，和一个字符，删除数组中的因此给定的字符，并进行压缩。M jia

压缩就是全变成_____, eg: 数组是  abcdeba, 给的字符是 b, 删除后数组变为acdea__,后面两个是空字符

void compressMe(char[] s, char c) {
    int index = 0;
    int i = 0;
    for (; i < s.length; i++){
        if (s[i] != c){
            s[index++] = s[i];
        }
    }
    for (; index < s.length; index++){
         s[index] = ' ';
    }
}

C++:

private static void compressMe(char[] s, char c)
        {
            int insertIdx = 0;
            for (int i = 0; i < s.Length; i++)
            {
                if (s[i] != c) s[insertIdx++] = s[i];
            }

            while (insertIdx < s.Length) s[insertIdx++] = ' ';
        }

Python:  一个循环里

class Solution(object):
    def moveZero(self, nums, elem):
        i = 0
        for j in xrange(len(nums)):
            if nums[j] != elem:
                nums[i], nums[j] = nums[j], ''
                i += 1
            else:
                nums[j] = ''

　

 

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