LeetCode第27題
- 2019 年 10 月 7 日
- 筆記
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]);}
翻譯：
給定一個數組和一個定值，刪除數組中的所有此值並返回一個新的數組長度
不要申明額外的數組空間，必須保證演算法複雜度為O(1)
數組的順序無所謂
思路：
這一題和【LeetCode第26題】差不多思路，只不過一個是刪除重複值，一個是刪除給定值
既然不能申明額外的數組，那隻能在原來的數組上做變動
變動前：[1,1,2,3,3]，給定值為：1
變動後：[2,3,3,3,3]
前3個值[2,3,3]就是我們所需要的
程式碼：
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; }}
因為最後一次進入if(nums[i] != val)判斷後，還執行了一次j++，所以j的值就已經等於數組長度了，return的時候不需要再+1了
