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题目:给定一个数组array和一个值value,移除掉数组中所有与value值相等的元素,返回新的数组的长度;要求:不能分配额外的数组空间,且必须使用原地排序的思想,空间复杂度O(1);
举例:
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.
解题思路:
1. 首先找到第一个等于value和第一个不等于value的数的位置;
2. 等于value和不等于value之间的数则全部为value;
3. 每次交换时一定是第一个等于value和第一个不等于value的数进行交换;
举例说明:
3,3,3,2,2,3,1,2,4,3,4,3,2 ; value = 3
1) 初始时:start = 0; index = 3; exchange两个位置的数,结果变为:2,3,3,3,2,3,1,2,4,3,4,3,2;
2) start = 1; index = 4; exchange: 2,2, 3,3,3,3,1,2,4,3,4,3,2;
3) start = 2; index = 5;因为nums[index] = 3,因此index一直递增,直到nums[index] != 3,此时index = 6; exchange: 2,2, 1,3,3,3,3,2,4,3,4,3,2;
4) start = 3; index = 7;exchange: 2,2, 1,2,3,3,3,3,4,3,4,3,2;
5) start = 4; index = 8; exchange: 2,2, 1,2,4,3,3,3,3,3,4,3,2;
6) start = 5; index = 9; nums[index] = 3, index递增,直到index = 10:exchange: 2,2, 1,2,4,4,3,3,3,3,3,3,2;
7) start = 6; index = 11; nums[index] = 3, index递增,直到index = 12: exchange: 2,2, 1,2,4,4,2,3,3,3,3,3,3;
此时start = 7;index = 12 等于数组长度,结束;
代码如下:
1 public class Solution { 2 public int removeElement(int[] nums, int val) { 3 if(nums == null || nums.length < 1){ 4 return 0; 5 } 6 int count = 0; 7 int index = 0; 8 while(index < nums.length && nums[index] != val){index ++; count ++;} //找到第一个等于value的数 9 int start = index; 10 while(index < nums.length && nums[index] == val){index ++;} //找到第一个不等于value的数 11 while(start < nums.length && index < nums.length){ 12 exchange(nums, start, index); 13 count ++; 14 start ++; 15 while(index < nums.length && nums[index] == val){index ++;} 16 17 } 18 return count; 19 } 20 public static void exchange(int[] nums, int index1, int index2) 21 { 22 int temp = nums[index1]; 23 nums[index1] = nums[index2]; 24 nums[index2] = temp; 25 } 26 }
Leetcode27--->Remove Element(移除数组中给定元素)
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原文地址:http://www.cnblogs.com/leavescy/p/5883198.html