【LeetCode】解题27：Remove Element

Problem 27: Remove Element [Easy]

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.

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 being modified to 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 = removeDuplicates(nums);

// 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]);
}

来源：LeetCode

解题思路

• 使用一个指针index指向需要修改的位置，另一个指针i指向要检测的数字。若nums[i]不等于val，将数字填入nums[index]，指针index向后移一位。

时间复杂度O(n)。

要点：双指针

Solution (Java)

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