Given a sorted array, remove the duplicates in place such that each element appear onlyonce and return the new length.
Do not allocate extra space for another array, you must do this in place with constant memory.
For example, Given input array nums = [1,1,2],
Your function should return length = 2, with the first two elements ofnums being1 and 2 respectively. It doesn't matter what you leave beyond the new length.
注意考虑清楚为啥循环else先pre++？？
最后为啥返回pre+1？？？
public int removeDuplicates(int[] A) { if (A.length <= 1) return A.length; int prev = 0; // point to previous int curr = 1; // point to current while (curr < A.length) { if (A[curr] == A[prev]) { curr++; } else { prev++; A[prev] = A[curr]; curr++; } } return prev + 1; }改进版
public int removeDuplicates(int[] nums) { if (nums == null || nums.length == 0) return 0; int i = 0; for (int n : nums) { // i<1代表第一个元素， // 当n>nums[i-1]代表不等，可以赋值，并且i++ if (i < 1 || n > nums[i - 1]) nums[i++] = n; } return i; }Follow up for "Remove Duplicates":
What if duplicates are allowed at most twice? Given sorted array nums = [1,1,1,2,2,3],
For example,
Your function should return length = 5, with the first five elements of nums being 1, 1, 2, 2 and 3. It doesn't matter what you leave beyond the new length.
同上一题的方法一
public int removeDuplicates(int[] A) { if (A.length <= 2) return A.length; int prev = 1; // point to previous int curr = 2; // point to current while (curr < A.length) { if (A[curr] == A[prev] && A[curr] == A[prev - 1]) { curr++; } else { prev++; A[prev] = A[curr]; curr++; } } return prev + 1; }
同上一题的方法二
public int removeDuplicates(int[] nums) { if (nums == null || nums.length == 0) return 0; int i = 0; for (int n : nums) { // //不同点 if (i < 2 || n > nums[i - 2]) nums[i++] = n; } return i; }
