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.
给定一个array和一个值，将数组中相同的值去掉，返回新的数组的长度。
算法分析
方法一：
算法跟LeetCode-26思路一样。具体参见Remove Duplicates from Sorted Array
代码
public class Solution {
public int removeElement(int[] nums, int val) {
int n = nums.length;
if (n == 0) return 0;
int index = 0;
for (int i = 0; i < n; i++) {
if (nums[i] != val) {
nums[index] = nums[i];
index ++;
}
}
return index;
}
}
方法二
现在考虑如下情况：nums=[1,2,3,5,4],val=4，方法一会复制前面四个值，很没有必要。因此可以考虑把这些没有必要的步骤去掉。如下情况：nums = [4,1,2,3,5], val = 4。nums[i] == val时，将左后一个值与当前值交换。
代码
public class Solution {
public int removeElement(int[] nums, int val) {
int n = nums.length;
int i = 0;
while (i < n) {
if (nums[i] == val) {
nums[i] = nums[n - 1];
n --;
} else {
i ++;
}
}
return i;
}
}
