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// 面试题21：调整数组顺序使奇数位于偶数前面 // 题目：输入一个整数数组，实现一个函数来调整该数组中数字的顺序，使得所有 // 奇数位于数组的前半部分，所有偶数位于数组的后半部分。 #include <cstdio> void Reorder(int *pData, unsigned int length, bool (*func)(int)); bool isEven(int n); // ====================方法一==================== void ReorderOddEven_1(int *pData, unsigned int length) { if (pData == nullptr || length == 0) return; int *pBegin = pData; int *pEnd = pData + length - 1; while (pBegin < pEnd) { // 向后移动pBegin，直到它指向偶数 while (pBegin < pEnd && (*pBegin & 0x1) != 0) pBegin++; // 向前移动pEnd，直到它指向奇数 while (pBegin < pEnd && (*pEnd & 0x1) == 0) pEnd--; if (pBegin < pEnd) { int temp = *pBegin; *pBegin = *pEnd; *pEnd = temp; } } } // ====================方法二==================== void ReorderOddEven_2(int *pData, unsigned int length) { Reorder(pData, length, isEven); } void Reorder(int *pData, unsigned int length, bool (*func)(int)) { if (pData == nullptr || length == 0) return; int *pBegin = pData; int *pEnd = pData + length - 1; while (pBegin < pEnd) { // 向后移动pBegin while (pBegin < pEnd && !func(*pBegin)) pBegin++; // 向前移动pEnd while (pBegin < pEnd && func(*pEnd)) pEnd--; if (pBegin < pEnd) { int temp = *pBegin; *pBegin = *pEnd; *pEnd = temp; } } } bool isEven(int n) { return (n & 1) == 0; } // ====================测试代码==================== void PrintArray(int numbers[], int length) { if (length < 0) return; for (int i = 0; i < length; ++i) printf("%d\t", numbers[i]); printf("\n"); } void Test(char *testName, int numbers[], int length) { if (testName != nullptr) printf("%s begins:\n", testName); int *copy = new int[length]; for (int i = 0; i < length; ++i) { copy[i] = numbers[i]; } printf("Test for solution 1:\n"); PrintArray(numbers, length); ReorderOddEven_1(numbers, length); PrintArray(numbers, length); printf("Test for solution 2:\n"); PrintArray(copy, length); ReorderOddEven_2(copy, length); PrintArray(copy, length); delete[] copy; } void Test1() { int numbers[] = {1, 2, 3, 4, 5, 6, 7}; Test("Test1", numbers, sizeof(numbers) / sizeof(int)); } void Test2() { int numbers[] = {2, 4, 6, 1, 3, 5, 7}; Test("Test2", numbers, sizeof(numbers) / sizeof(int)); } void Test3() { int numbers[] = {1, 3, 5, 7, 2, 4, 6}; Test("Test3", numbers, sizeof(numbers) / sizeof(int)); } void Test4() { int numbers[] = {1}; Test("Test4", numbers, sizeof(numbers) / sizeof(int)); } void Test5() { int numbers[] = {2}; Test("Test5", numbers, sizeof(numbers) / sizeof(int)); } void Test6() { Test("Test6", nullptr, 0); } int main(int argc, char *argv[]) { Test1(); Test2(); Test3(); Test4(); Test5(); Test6(); return 0; } // ------ Output ------ /* Test1 begins: Test for solution 1: 1 2 3 4 5 6 7 1 7 3 5 4 6 2 Test for solution 2: 1 2 3 4 5 6 7 1 7 3 5 4 6 2 Test2 begins: Test for solution 1: 2 4 6 1 3 5 7 7 5 3 1 6 4 2 Test for solution 2: 2 4 6 1 3 5 7 7 5 3 1 6 4 2 Test3 begins: Test for solution 1: 1 3 5 7 2 4 6 1 3 5 7 2 4 6 Test for solution 2: 1 3 5 7 2 4 6 1 3 5 7 2 4 6 Test4 begins: Test for solution 1: 1 1 Test for solution 2: 1 1 Test5 begins: Test for solution 1: 2 2 Test for solution 2: 2 2 Test6 begins: Test for solution 1: Test for solution 2: */
