【LeetCode算法-27】Remove Element

LeetCode第27题

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]); }

翻译：

给定一个数组和一个定值，删除数组中的全部此值并返回一个新的数组长度

不要申明额外的数组空间，必须保证算法复杂度为O(1)

数组的顺序无所谓

思路：

这一题和【LeetCode算法-26】Remove Duplicates from Sorted Array差很少思路，只不过一个是删除重复值，一个是删除定值

既然不能申明额外的数组，那只能在原来的数组上作变更

变更前：[1,1,2,3,3]，定值为：1

变更后：[2,3,3,3,3]

前3个值[2,3,3]就是咱们所须要的

代码：

class Solution { public int removeElement(int[] nums, int val) { if(nums == null || nums.length == 0) return 0; int j = 0; for(int i = 0;i<nums.length;i++){ if(nums[i] != val){ nums[j] = nums[i]; System.out.println(nums[j]); j++; } } return j; } }

由于最后一次进入if(nums[i] != val)判断后，仍是执行了一次j++，因此j的值就已是数组长度了，return的时候不须要再+1了

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