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| 1 | +/* |
| 2 | + * Copyright (C) 2022, Saul Lawliet <october dot sunbathe at gmail dot com> |
| 3 | + * All rights reserved. |
| 4 | + * |
| 5 | + * Follow up: |
| 6 | + * - It is very easy to come up with a solution with a runtime of O(n log n). Can you do it in linear time O(n) and possibly in a single pass? |
| 7 | + * |
| 8 | + * 如果是 O(n) 的时间复杂度, 那么思考方向应该类似于动态规划的做法 |
| 9 | + * PS: 尝试引入变量, 来减少循环中的计算, 但效果不明显, 可能原因是: 相同的代码每次执行的时候可能会有不同的运行时间 |
| 10 | + */ |
| 11 | + |
| 12 | +#include "c/data-structures/array.h" |
| 13 | +#include "c/test.h" |
| 14 | + |
| 15 | +/** |
| 16 | + * Note: The returned array must be malloced, assume caller calls free(). |
| 17 | + */ |
| 18 | +int *countBits(int n, int *returnSize) { |
| 19 | + *returnSize = n + 1; |
| 20 | + int *returnArray = malloc(sizeof(int) * *returnSize); |
| 21 | + returnArray[0] = 0; |
| 22 | + if (n >= 1) returnArray[1] = 1; |
| 23 | + int pow = 4; |
| 24 | + for (int i = 2; i < *returnSize; i++) { |
| 25 | + if (i < pow) { |
| 26 | + returnArray[i] = 1 + returnArray[i - pow / 2]; |
| 27 | + } else { |
| 28 | + returnArray[i] = 1; |
| 29 | + pow <<= 1; |
| 30 | + } |
| 31 | + } |
| 32 | + return returnArray; |
| 33 | +} |
| 34 | + |
| 35 | +void test(const char *expect, int n) { |
| 36 | + int returnSize; |
| 37 | + int *returnArray = countBits(n, &returnSize); |
| 38 | + |
| 39 | + EXPECT_EQ_STRING_AND_FREE_ACTUAL(expect, arrayToString1D(returnArray, returnSize, ARRAY_INT)); |
| 40 | +} |
| 41 | + |
| 42 | +int main(void) { |
| 43 | + test("[0,1,1]", 2); |
| 44 | + test("[0,1,1,2,1,2]", 5); |
| 45 | + |
| 46 | + return testOutput(); |
| 47 | +} |
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