leetCode- Remove Element
阿新 • • 發佈:2017-11-08
ray this .org remove 情況 blog get for order
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 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:
Given nums = [3,2,2,3], val = 3, Your function should return length = 2, with the first two elements of nums being 2.
我的版本(適合刪除元素很多的情況):
class Solution { public int removeElement(int[] nums, int val) { int count=0; for(int i=0;i<nums.length;i++){ if(nums[i]!=val){ nums[count++] = nums[i]; } }return count; } }
其他版本(適合刪除元素很少的情況):
public int removeElement(int[] nums, int val) {
int i = 0;
int n = nums.length;
while (i < n) {
if (nums[i] == val) {
nums[i] = nums[n - 1];
// reduce array size by one
n--;
} else {
i++;
}
}
return n;
}
leetCode- Remove Element