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 ofnumsbeing1and2respectively. 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 ofnumsbeing modified to0,1,2,3, and4respectively. 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]);
}
//C++
class Solution {
public:
int removeDuplicates(vector<int>& nums) {
int l = nums.size();
if(l == 0) return 0;
int index = 0;
for(int i = 1; i < l; i++) {
if(nums[i] != nums[index])
nums[++index] = nums[i];
}
return index + 1;
}
};
本文介绍了一种在不使用额外空间的情况下,修改输入数组以去除重复元素的方法。通过一个指针跟踪唯一元素的位置,当遇到不同的元素时,将其移动到该位置并更新指针,最终返回新的长度。这种方法仅使用常数级的额外内存。
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