{

"files.associations": {

"ostream": "cpp"

}

}

int KZero(const vector<int>& nums){

int max_length = 0, left = 0, right = 0, num = 0, sz = nums.size();

while(right < sz){

if(nums[right] == 0){

//更新left到第一个非0序列

while(left < sz && nums[left] == 0){

left++;

}

}

right = left;

if(right > sz) break;

while(nums[right] == 1) right++;

max_length = max(max_length, right - left);

}

return max_length;

}

int longestOnes(vector<int>& nums, int k) {

if(k == 0) return KZero(nums);

//这道题，我们可以更改思路，因为翻转k个0并不好算，所以我们可以找连续k个0的个数，找到了就是结果

int max_length = 0, left = 0, right = 0, num = 0, sz = nums.size();

while(right < sz){

if(num >= k){

//证明[left, right)区间正好有k个0，此时要将left指向最近的0的下一个位置

while(left < right && nums[left] != 0) { left++; }

left++, num--;

}

while(left < right && nums[right] != 0) { right++; }

right++, num++;

while(right < sz && nums[right] != 0) { right++; }

max_length = max(max_length, right-left);

cout << max_length << "left is " << left << "right is " << right << endl;

return max_length;

vector<int> nums = {0,0,1,1,1,0,0};

cout << s.longestOnes(nums, 0) << endl;

test: test.cpp

g++ test.cpp -o test

PHONY: clean

clean:

rm test

#include <iostream>

using namespace std;

int main(){

int x = int();

char m = char();

double d = double();

cout << m << endl;

cout << x << " " << m << " " << d << endl;

return 0;

}