zoukankan      html  css  js  c++  java
  • 26. Remove Duplicates from Sorted Array

    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:

    AC code:

    class Solution {
    public:
        int removeDuplicates(vector<int>& nums) {
            int len = nums.size();
            int res = len;
            if (len == 0) return 0;
            vector<int> v;
            v.push_back(nums[0]);
            for (int i = 1; i < len; ++i) {
                if (nums[i] == nums[i-1]) {
                    res--;
                } else {
                    v.push_back(nums[i]);
                }
            }
            nums = v;
            return res;
        }
    };
    
    Runtime: 36 ms, faster than 23.42% of C++ online submissions for Remove Duplicates from Sorted Array.
    永远渴望,大智若愚(stay hungry, stay foolish)
  • 相关阅读:
    router-link中传值的三种方式
    JVM原理和优化
    JAVA中关于锁机制
    思考程序
    论防御式编程与攻击式编程
    BOM详解
    理解JAVASCRIPT 闭包
    用HTML5 CANVAS做自定义路径的动态效果图片!
    js制作点击会自动隐藏的导航栏(固定在在头部的)
    ++a和a++的区别。
  • 原文地址:https://www.cnblogs.com/h-hkai/p/9748499.html
Copyright © 2011-2022 走看看