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.
分析一下：
条件：数组，排好序
要求：1、删除所有重复的数字，每种只留一个；
2、把不重复的放前面
3、space = O(1)
要点：利用好条件 1、排好序的数组
2、分块操作？
3、加入指针操作？
代码
package leetcode;
public class RemoveDuplicatesfromSortedArray {
public int removeDuplicates(int[] nums) {
int m = nums.length;
int index = 1;
if(m==1) return 1;
for(int i=0;i<m-1;i++){ //就是判断前后两个元素的大小，不相同就转移，相同继续，注意越界问题；
if(nums[i]<nums[i+1]){
nums[index++] = nums[i+1]; //每次换完就自++；
}
}
return index;
}
public static void main(String[] args) {
// TODO Auto-generated method stub
}
}