Given a sorted array, 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 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 of nums being 1 and 2 respectively. It doesn't matter what you leave beyond the new length. ~~~ public class Solution { public int removeDuplicates(int[] A) { if(A.length ==0){ return 0; } int count=1; for(int i=1; i<A.length; ++i){ if(A[i] != A[i-1]){ //注意這行代碼 A[count]=A[i]; count++; } } return count; } } // 若數族大小為0/1。 數組A 不變，返回0/1。 // 若數組大小大于1。用count記錄新數組元素A的末尾的下一個index，若a[i] != a[i-1], 則把 a[i]的值賦給count位置上（a[count] = a[i] )。 // 例如數組元素為 1, 2 2, 2 2, 3, 3, 4。 此時count=2,表示A的下標2前面有兩個不同的元素，若 A[5]=3,A[4]=2, 則到A[5] != A[4]時，將A[5]的只賦給A[count]上，然后count++ ~~~