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| 1 | +/* |
| 2 | + * @Author: Chacha |
| 3 | + * @Date: 2022-07-24 22:29:36 |
| 4 | + * @Last Modified by: Chacha |
| 5 | + * @Last Modified time: 2022-07-24 23:08:35 |
| 6 | + */ |
| 7 | + |
| 8 | +/** |
| 9 | + * 来源:https://leetcode.cn/problems/squares-of-a-sorted-array/ |
| 10 | + * |
| 11 | + * 977. 有序数组的平方 |
| 12 | + * |
| 13 | + * 给你一个按 非递减顺序 排序的整数数组 nums,返回 每个数字的平方 组成的新数组,要求也按 非递减顺序 排序。 |
| 14 | + * |
| 15 | + * 示例 1: |
| 16 | + * 输入:nums = [-4,-1,0,3,10] |
| 17 | + * 输出:[0,1,9,16,100] |
| 18 | + * 解释:平方后,数组变为 [16,1,0,9,100],排序后,数组变为 [0,1,9,16,100] |
| 19 | + * |
| 20 | + * 示例 2: |
| 21 | + * 输入:nums = [-7,-3,2,3,11] |
| 22 | + * 输出:[4,9,9,49,121] |
| 23 | + * |
| 24 | + * 提示: |
| 25 | + * 1. 1 <= tokens.length <= 10^4 |
| 26 | + * 2. -10^4 <= nums[i] <= 10^4 |
| 27 | + * 3. nums 已按 非递减顺序 排序 |
| 28 | + * |
| 29 | + */ |
| 30 | + |
| 31 | +#include <iostream> |
| 32 | +#include <vector> |
| 33 | + |
| 34 | +using namespace std; |
| 35 | + |
| 36 | +class Solution |
| 37 | +{ |
| 38 | +private: |
| 39 | + /* data */ |
| 40 | +public: |
| 41 | + vector<int> sortSquares(vector<int> &nums); |
| 42 | +}; |
| 43 | + |
| 44 | +/** |
| 45 | + * 方法:双指针法 |
| 46 | + * |
| 47 | + * 数组其实是有序的,只不过负数平方之后可能成为最大数了,那么数组平方的最大值就在数组的两端,不是最左边就是最右边,不可能是中间。 |
| 48 | + * 此时可以考虑双指针法了,i指向起始位置,j指向终止位置。定义一个新数组result,和A数组一样的大小,让k指向result数组终止位置。 |
| 49 | + * 如果A[i] * A[i] < A[j] * A[j] 那么result[k--] = A[j] * A[j]; |
| 50 | + * 如果A[i] * A[i] >= A[j] * A[j] 那么result[k--] = A[i] * A[i]; |
| 51 | + * |
| 52 | + */ |
| 53 | +vector<int> Solution::sortSquares(vector<int> &nums) |
| 54 | +{ |
| 55 | + int k = nums.size() - 1; |
| 56 | + vector<int> result(nums.size(), 0); |
| 57 | + |
| 58 | + for (int i = 0, j = nums.size() - 1; i <= j;) |
| 59 | + { |
| 60 | + // 注意这里要i <= j,因为最后要处理两个元素 |
| 61 | + if (nums[i] * nums[i] < nums[j] * nums[j]) |
| 62 | + { |
| 63 | + result[k--] = nums[j] * nums[j]; |
| 64 | + j--; |
| 65 | + } |
| 66 | + else |
| 67 | + { |
| 68 | + result[k--] = nums[i] * nums[i]; |
| 69 | + i++; |
| 70 | + } |
| 71 | + } |
| 72 | + |
| 73 | + return result; |
| 74 | +} |
| 75 | + |
| 76 | +void printVector(vector<int> &vec) |
| 77 | +{ |
| 78 | + for (int i = 0; i < vec.size(); i++) |
| 79 | + { |
| 80 | + printf("%3d ", vec[i]); |
| 81 | + } |
| 82 | + cout << endl; |
| 83 | +} |
| 84 | + |
| 85 | +int main(int argc, char const *argv[]) |
| 86 | +{ |
| 87 | + Solution s; |
| 88 | + |
| 89 | + vector<int> nums = {-4, -1, 0, 3, 10}; |
| 90 | + vector<int> _nums = s.sortSquares(nums); |
| 91 | + |
| 92 | + cout << "nums = [-4, -1, 0, 3, 10], 运算结果为: " << endl; |
| 93 | + |
| 94 | + printVector(_nums); |
| 95 | + |
| 96 | + return 0; |
| 97 | +} |
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