Given an array and a value, 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 in place with constant memory.

The order of elements can be changed. It doesn't matter what you leave beyond the new length.

Example:

Given input array nums = [3,2,2,3], val = 3

Your function should return length = 2, with the first two elements of nums being 2.

 

Hint:

1. Try two pointers.
2. Did you use the property of "the order of elements can be changed"?
3. What happens when the elements to remove are rare?

 

　　题目不许开辟新空间来定义新数组,所以就只能在所给的数组中解决问题.

　　题目中说,The order of elements can be changed.通过改变元素的顺序,来完成.

　　思路: 定义一个计数器count,从零开始.每遇到一个非指定数字(val),就把它赋值给从零开始的元素,每赋值一次,count++,则遍历完所有数组元素之后,得到的count就是含有除了val以外的元素的数组的长度,原数组的count长度,就是新数组.

1 int removeElement(int* nums, int numsSize, int val) { 2     int j = 0; 3     for(int i = 0; i < numsSize; i++){ 4         if(nums[i] != val){ 5             nums[j] = nums[i]; 6             j++; 7         } 8     } 9     return j;10 }

 

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