LeetCode-27. Remove Element(Java)
2017-07-13 15:00
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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 =
Your function should return length = 2, with the first two elements of nums being 2.
-------------------------------------------------------------------------------------------------------------------------------------------------------------
题意
删除数组中指定元素,并返回删除后的数组长度,要求不能用额外的空间,元素的顺序可以被改变。
思路
首先遇到与给定元素相同的,length减1,由于不考虑元素顺序,则j将元素相同的位置替换成其他的就可以了。
代码
public class Solution {
public int removeElement(int[] nums, int val) {
int m = 0;
for(int i = 0; i < nums.length; i++){
if(nums[i] != val){
nums[m] = nums[i];
m++;
}
}
return m;
}
}本方法借助原来的数组,将与给定元素不相同的元素覆盖到m位置上。在删除元素的同时,元素顺序也保持不变。
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.
-------------------------------------------------------------------------------------------------------------------------------------------------------------
题意
删除数组中指定元素,并返回删除后的数组长度,要求不能用额外的空间,元素的顺序可以被改变。
思路
首先遇到与给定元素相同的,length减1,由于不考虑元素顺序,则j将元素相同的位置替换成其他的就可以了。
代码
public class Solution {
public int removeElement(int[] nums, int val) {
int m = 0;
for(int i = 0; i < nums.length; i++){
if(nums[i] != val){
nums[m] = nums[i];
m++;
}
}
return m;
}
}本方法借助原来的数组,将与给定元素不相同的元素覆盖到m位置上。在删除元素的同时,元素顺序也保持不变。
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