题目:
Given a sorted array nums, remove the duplicates in-place such that each element appear only once 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.
Example 1:
Given nums = [1,1,2], Your function should return length = 2, with the first two elements of nums being 1 and 2 respectively. It doesn't matter what you leave beyond the returned length.
Example 2:
Given nums = [0,0,1,1,1,2,2,3,3,4], Your function should return length = 5, with the first five elements of nums being modified to 0, 1, 2, 3, and 4 respectively. 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 = removeDuplicates(nums);
// 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]); }
public class RemoveDuplicatesSortedArray {
public static int removeDuplicates(int[] nums) {
if (nums == null)
return 0;
if (nums.length < 2)
return nums.length;
int len=0;
int i =0;
int j = 1;
while(i<nums.length-1 && j<nums.length) {
if(nums[i]!=nums[j]) {
nums[i+1]=nums[j];
i++;
}
j++;
len = i+1;
}
return len;
}
public static void main(String[] args) {
int[] nums = {0,0,1,1,1,2,2,3,3,4};
System.out.println(removeDuplicates(nums));
}
}
博客给出一道LeetCode题目,要求在排序数组中原地移除重复项,使每个元素仅出现一次,并返回新长度,且只能使用O(1)额外内存。还给出两个示例及对返回值为整数但答案为数组的解释。
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