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.
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.
题意：
给定某个值，要求删除数组中所有等于该值的元素。
思路：
j
以[3, 2, 2, 3]， value = 3 为例
i
指针i遍历数组，遇到非value的值，送到指针j那里去
指针j从0开始，接收所有指针i送来的非value值
扫完一遍后
指针j所指的位置就是新“数组”的长度
代码：
1 public int removeElement(int[] nums, int target) { 2 int index = 0; 3 for (int i = 0; i < nums.length; ++i) { 4 if (nums[i] != target) { 5 nums[index++] = nums[i]; 6 } 7 } 8 return index; 9 }
转载于:https://www.cnblogs.com/liuliu5151/p/9125368.html
