LeetCode 27. Remove Element(java)

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 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:

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

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

我们用induction的方法来想，假设0 ~ i-1成立，并且返回值为len，那么我们看从i-1到i的情况(i >= 1)：1. 如果nums[i] == val, do nothing; 2. 如果nums[i] != val, nums[len] = nums[i], len++. 理清了递归关系，接下来我们来看base case: i = 0的时候，len = nums[0] == 3 ? 0 : 1. 最后就是special case，我们判断一下就可以了。最后的最后，我们需要判断base case能不能写进general case里，如果可以，那么久代码合并。

public int removeElement(int[] nums, int val) {
//special case
if (nums == null || nums.length == 0) return 0;
//base case
int len = 0;
//general case
for (int i = 0; i < nums.length; i++) {
if (nums[i] != val) {
nums[len] = nums[i];
len++;
}
}
return len;
}