[LeetCode] 360. Sort Transformed Array 排序转换后的数组

Given a sorted array of integers nums and integer values a, b and c. Apply a function of the form f(x) = ax² + bx + c to each element x in the array.

The returned array must be in sorted order.

Expected time complexity: O(n)

Example:

nums = [-4, -2, 2, 4], a = 1, b = 3, c = 5,

Result: [3, 9, 15, 33]

nums = [-4, -2, 2, 4], a = -1, b = 3, c = 5

Result: [-23, -5, 1, 7]

Credits:
Special thanks to @elmirap for adding this problem and creating all test cases.

解法1：把每一个数带入方程计算结果，然后对结果排序。Time: O(nlogn)，不是题目想要的

解法2：根据抛物线方程式的特点，a>0则抛物线开口朝上，两端的值比中间的大，a<0则抛物线开口朝下，则两端的值比中间的小，a=0时，则为直线方法，是单调递增或递减的。同时因为给定数组nums是有序的，所以才有O(n)的解法。

当a>0，两端的值比中间的大，从后边往中间填数，用两个指针分别指向数组的开头和结尾，比较两个数，将其中较大的数存入res，然后该指针向中间移，重复比较过程，直到把res都填满。

当a<0，两端的值比中间的小，从前边往中间填数，用两个指针分别指向数组的开头和结尾，比较两个数，将其中较小的数存入res，然后该指针向中间移，重复比较过程，直到把res都填满。

当a=0，函数是单调递增或递减的，那么从前往后填和从后往前填都可以，也可以将这种情况和a>0合并。

Java:

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public class Solution {        private int calcu(int x, int a, int b, int c){         return a*x*x + b*x + c;     }                 public int[] sortTransformedArray(int[] nums, int a, int b, int c) {         int index;         if(a > 0){             index = nums.length - 1 ;         } else {             index = 0;         }         int result[] = new int[nums.length];         int i = 0;         int j = nums.length - 1;         if(a > 0){             while(i <= j){                 result[index--] = calcu(nums[i],a,b,c) > calcu(nums[j],a,b,c) ? calcu(nums[i++],a,b,c):calcu(nums[j--],a,b,c);             }         }else{             while(i <= j){                 result[index++] = calcu(nums[i],a,b,c) < calcu(nums[j],a,b,c) ? calcu(nums[i++],a,b,c):calcu(nums[j--],a,b,c);             }          }         return result;     } }

Python: wo

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class Solution():     def sortTransformedArray(self, nums, a, b, c):         if not nums:             return []         res = [0] * len(nums)         i, j = 0, len(nums) - 1         index = len(nums) - 1 if a >= 0 else 0         while i <= j:             if a >= 0:                 calc_i = self.calc(nums[i], a, b, c)                 calc_j = self.calc(nums[j], a, b, c)                 if  calc_i >= calc_j:                     res[index] = calc_i                     i += 1                 else:                     res[index] = calc_j                     j -= 1                 index -= 1             else:                 calc_i = self.calc(nums[i], a, b, c)                 calc_j = self.calc(nums[j], a, b, c)                 if  calc_i <= calc_j:                     res[index] = calc_i                     i += 1                 else:                     res[index] = calc_j                     j -= 1                 index += 1           return res       def calc(self, x, a, b, c):         return  a * x * x + b * x + c     if __name__ == '__main__':     print Solution().sortTransformedArray([], 1, 3, 5)     print Solution().sortTransformedArray([2], 1, 3, 5)     print Solution().sortTransformedArray([-4, -2, 2, 4], 1, 3, 5)     print Solution().sortTransformedArray([-4, -2, 2, 4], -1, 3, 5)     print Solution().sortTransformedArray([-4, -2, 2, 4], 0, 3, 5)

C++:

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class Solution { public:     vector<int> sortTransformedArray(vector<int>& nums, int a, int b, int c) {         int n = nums.size(), i = 0, j = n - 1;         vector<int> res(n);         int idx = a >= 0 ? n - 1 : 0;         while (i <= j) {             if (a >= 0) {                 res[idx--] = cal(nums[i], a, b, c) >= cal(nums[j], a, b, c) ? cal(nums[i++], a, b, c) : cal(nums[j--], a, b, c);             } else {                 res[idx++] = cal(nums[i], a, b, c) >= cal(nums[j], a, b, c) ? cal(nums[j--], a, b, c) : cal(nums[i++], a, b, c);             }         }         return res;     }     int cal(int x, int a, int b, int c) {         return a * x * x + b * x + c;     } };

C++:

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class Solution { public:     vector<int> sortTransformedArray(vector<int>& nums, int a, int b, int c) {         if(nums.size() ==0) return {};         vector<int> result;         int left = 0, right = nums.size()-1;         auto func = [=](int x) { return a*x*x + b*x + c; };         while(left <= right)         {             int val1 = func(nums[left]), val2 = func(nums[right]);             if(a > 0) result.push_back(val1>=val2?val1:val2);             if(a > 0) val1>val2?left++:right--;             if(a <= 0) result.push_back(val1>=val2?val2:val1);             if(a <= 0) val1>val2?right--:left++;         }         if(a > 0) reverse(result.begin(), result.end());         return result;     } };

　　

　　

　

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