算法作业HW26：LeetCode 27. Remove Element

Description:

Given an array and a value, remove all instances of that value in place and return the new length.

Do not allocate extra space for another array, you must do this in place with constant memory.

The order of elements can be changed. It doesn't matter what you leave beyond the new length.

 

Note:

Example:
Given input array nums = [3,2,2,3], val = 3

Your function should return length = 2, with the first two elements of nums being 2.

Solution:

  Analysis and Thinking:

给定一个数组以及一个目标值，移除数组中所有与该值相等的元素，并返回移除后数组长度。与26一样，通过理解题意以及输入数组长度，设置循环即可解决

  Steps:

1. 设置记录变量count

2. 遍历数组，如果当前值与目标值相等，count加一，跳过，相当于移除

3. 如果当前值与目标值不等，设置当前位置减去count下标的位置的元素为当前元素，

4. 不断重复，最后构造了一个新的不包含目标值的数组

5  返回数组长度减去count的值，即结果，因为count代表了等于目标值的元素个数

Codes:

class Solution {public:    int removeElement(vector<int>& nums, int val) {    int count = 0;    for(int i = 0 ; i < nums.size() ; ++i) {        if(nums[i] == val)            count++;        else            nums[i-count] = nums[i];    }

    int result = nums.size()-count;    return result;}};

Results: