题目:
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]);
}
public class RemoveElement {
public static int removeElement(int[] nums, int val) {
int len = nums.length;
if (nums == null || nums.length == 0)
return 0;
int k = nums.length - 1;
int i = 0;
while(i <= k){
if (nums[i] == val){
swap(nums, i, k--);
len--;
}else {
i++;
}
}
return len;
}
public static void swap(int arr[], int i, int j){
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
public static void main(String[] args) {
//int[] nums = {0,1,2,2,3,0,4,2};
int[] nums = {3,2,2,3};
//int val = 2;
int val = 3;
System.out.println(removeElement(nums, val));
}
}
博客围绕LeetCode题目展开,要求在不额外分配数组空间的情况下,原地移除数组中指定值并返回新长度。给出了两个示例,还解释了返回值为整数但答案涉及数组的原因,强调输入数组是按引用传递,函数内对数组的修改调用者可知。
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