Given two integers tomatoSlices and cheeseSlices. The ingredients of different burgers are as follows:
- Jumbo Burger: 4 tomato slices and 1 cheese slice.
- Small Burger: 2 Tomato slices and 1 cheese slice.
Return [total_jumbo, total_small] so that the number of remaining tomatoSlices equal to 0 and the number of remaining cheeseSlices equal to 0. If it is not possible to make the remaining tomatoSlices and cheeseSlices equal to 0 return [].
Example 1:
Input: tomatoSlices = 16, cheeseSlices = 7 Output: [1,6] Explantion: To make one jumbo burger and 6 small burgers we need 4*1 + 2*6 = 16 tomato and 1 + 6 = 7 cheese. There will be no remaining ingredients.
Example 2:
Input: tomatoSlices = 17, cheeseSlices = 4 Output: [] Explantion: There will be no way to use all ingredients to make small and jumbo burgers.
Example 3:
Input: tomatoSlices = 4, cheeseSlices = 17 Output: [] Explantion: Making 1 jumbo burger there will be 16 cheese remaining and making 2 small burgers there will be 15 cheese remaining.
Example 4:
Input: tomatoSlices = 0, cheeseSlices = 0 Output: [0,0]
Example 5:
Input: tomatoSlices = 2, cheeseSlices = 1 Output: [0,1]
Constraints:
0 <= tomatoSlices <= 10^70 <= cheeseSlices <= 10^7
class Solution { public List<Integer> numOfBurgers(int tomatoSlices, int cheeseSlices) { List<Integer> res = new ArrayList(); if(tomatoSlices == 0 && cheeseSlices == 0){ res.add(0); res.add(0); return res; } int half = cheeseSlices / 2; //int[][] arr = new int[half + 1][2]; for(int i = 0; i <= half; i++){ int t1 = i; int t2 = cheeseSlices - i; if(t1 * 4 + t2 * 2 == tomatoSlices){ res.add(t1); res.add(t2); } else if(t1 * 2 + t2 * 4 == tomatoSlices){ res.add(t2); res.add(t1); } } return res; } }
CNM, O（n）都给TLE了
看了下解答，竟然是鸡兔同笼问题，一想确实有道理
class Solution { public List<Integer> numOfBurgers(int t, int c) { List<Integer> res = new ArrayList(); if(t % 2 == 0 && c * 2 <= t && t <= c * 4){ res.add(t/2 - c); res.add(c * 2 - t / 2); } else return res; return res; } }
