【双指针】便于线性表操作

Leetcode: 27
/Description:Given an array nums and a value val, 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 by modifying the input array in-place with O(1) extra memory.
The order of elements can be changed. It doesn’t matter what you leave beyond the new 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 a modification to the input array will be known to the caller as well.
Internally you can think of this:// nums is passed in by reference. (i.e., without making a copy)int len = removeElement(nums, val);
// any modification to nums in your function would be known by the caller.// using the length returned by your function, it prints the first len elements.for (int i = 0; i < len; i++) { print(nums[i]);}
Constraints:0 <= nums.length <= 1000 <= nums[i] <= 500 <= val <= 100/#includeusing namespace std;/思路：如何不按照顺序移除元素？双指针yyds/class Solution {public: int removeElement(vector& nums, int val) { int fastPtr = 0, slowPtr = 0; for (;fastPtr < nums.size(); fastPtr++) if (nums[fastPtr] != val) nums[slowPtr++] = nums[fastPtr]; return slowPtr; }};
int main() { vector nums = {3,2,2,3}; int val = 3; Solution ss; cout << ss.removeElement(nums, val) << endl; return 0;}