【本文链接】
http://www.cnblogs.com/hellogiser/p/binary-search-for-repeated-element.html
【题目】
给定一个升序排列的自然数数组，数组中包含重复数字，例如：[1,2,2,3,4,4,4,5,6,7,7]。问题：给定任意自然数，对数组进行二分查找，返回数组正确的位置，给出函数实现。注：连续相同的数字，返回第一个匹配位置还是最后一个匹配位置，由函数传入参数决定。
【代码】
C++ Code
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/*
version: 1.0 author: hellogiser blog: http://www.cnblogs.com/hellogiser date: 2014/9/22 */ #include "stdafx.h" #include "iostream" using namespace std; enum MATCH_POS { FIRST, LAST }; int binary_search(int *a, int n, int value) { if(a == NULL || n <= 0) return -1; int low = 0, high = n - 1, mid; while(low <= high) { mid = low + (high - low) / 2; if (value == a[mid]) { return mid; } else if (value < a[mid]) { high = mid - 1; } else { low = mid + 1; } } return -1; } int binary_search_2(int *a, int n, int value, MATCH_POS match) { if(a == NULL || n <= 0) return -1; int low = 0, high = n - 1, mid; //========================== int count = 0; //========================== while(low <= high) { //========================== count ++; //========================== mid = low + (high - low) / 2; if (value == a[mid]) { if (match == FIRST) { if (mid > 0 && a[mid - 1] == a[mid]) { high = mid - 1; } else { //========================== cout << "count = " << count << endl; //========================== return mid; } } else if (match == LAST) { if (mid < n - 1 && a[mid + 1] == a[mid]) { low = mid + 1; } else { //========================== cout << "count = " << count << endl; //========================== return mid; } } } else if (value < a[mid]) { high = mid - 1; } else { low = mid + 1; } } return -1; } void test_base2() { int a[] = {1, 2, 2, 3, 4, 4, 4, 5, 6, 7, 7}; for (int i = 0; i < sizeof(a) / 4; i++) { cout << binary_search_2(a, sizeof(a) / 4, a[i], FIRST) << endl; cout << binary_search_2(a, sizeof(a) / 4, a[i], LAST) << endl; } cout << "=================================" << endl; int a2[] = {1, 2, 3, 4, 5, 6, 7, 8, 9}; for (int i = 0; i < sizeof(a2) / 4; i++) { cout << binary_search_2(a2, sizeof(a2) / 4, a2[i], FIRST) << endl; cout << binary_search_2(a2, sizeof(a2) / 4, a2[i], LAST) << endl; } cout << "=================================" << endl; int a3[1024] = {0}; cout << binary_search_2(a3, sizeof(a3) / 4, 0, FIRST) << endl; cout << binary_search_2(a3, sizeof(a3) / 4, 0, LAST) << endl; } void test_main() { test_base2(); } int main(void) { test_main(); return 0; } |
由此可知，如果数组有1024个重复元素0，那么查找任然是只需要10次即可。
【参考】