一道面试题

《程序员面试宝典》14.5 字符串其它问题 面试例题 6：（Page 217）

有一个数组a[n]，里面的数只有两种：-1或1。i、j是两个整数，假设0 <= i <= j <= n-1，找出a[i]到a[j]中连续数之和最大的部分（如果最大部分存在相等的，则优先找最短的）。

感觉原文中给出的解法不正确，于是自己实现了一段代码。

思路：从i = 0至n-1，依次求出以a[i]为结尾的序列的最大和maxSum[i]，max ( maxSum[i] )即为最大和，其i值就是我们要求的end。同理从i=n-1至0，求一遍最大和，其下标即是我们要求的begin。考虑到会有多个区间符合要求，最后还要对区间跨度进行比较，选择一个跨度最小的。下面是代码：

#include <iostream>using namespace std;//arr :input array//len :array lenth//begin :return the begin index//end : return the end index//function return: max sumint find_range (int *arr, int len, int &begin, int &end){        int *maxSum;        int *begins;        int *ends;        maxSum = (int*)malloc (sizeof(int) * len);        //find the ends of the ranges witch has the max sum.        maxSum[0] = arr[0];        int max = maxSum[0];        for (int i = 1; i < len; i++) {                if (maxSum[i - 1] <= 0)                        maxSum[i] = arr[i];                else                        maxSum[i] = maxSum[i - 1] + arr[i];                if (maxSum[i] > max) {                        max = maxSum[i];                }        }        int ends_num = 0;        for (int i = 0; i < len; i++) {                if (maxSum[i] == max)                        ends_num++;        }        ends = (int*)malloc (sizeof (int) * ends_num);        for (int i = 0, j = 0; i < len; i++) {                if (maxSum[i] == max)                        ends[j++] = i;        }        //find the begins of the ranges witch has the max sum.        maxSum[len - 1] = arr[len - 1];        max = maxSum[len - 1];        for (int i = len - 2; i >= 0; i--) {                if (maxSum[i + 1] <= 0)                        maxSum[i] = arr[i];                else                        maxSum[i] = maxSum[i + 1] + arr[i];                if (maxSum[i] > max) {                        max = maxSum[i];                }        }        int begins_num = 0;        for (int i = 0; i < len; i++) {                if (maxSum[i] == max)                        begins_num++;        }        begins = (int*)malloc (sizeof (int) * begins_num);        for (int i = 0, j = 0; i < len; i++) {                if (maxSum[i] == max)                        begins[j++] = i;        }        //find the minum range [begin, end] having the max sum.        begin = begins[0];        end = ends[0];        int begin_i = 0, end_i = 0;        int step = end - begin;        while (begin_i < begins_num && end_i < ends_num) {                while ((begin_i + 1) < begins_num && begins[begin_i+1] <= ends[end_i])                        begin_i++;                if (step > (ends[end_i] - begins[begin_i]))                {                        begin = begins[begin_i];                        end = ends[end_i];                        step = end -begin;                }                if ((begin_i + 1) == begins_num)                        break;                else                        begin_i++;                while ((end_i + 1) < ends_num && ends[end_i + 1] < begins[begin_i])                        end_i++;        }        cout << "Max Sum = " << max << endl;        cout << "begin = " << begin << endl;        cout << "end = " << end << endl;        cout << "step = " << step << endl;        return max;}int main (){        //Test 1:        int arr1[10] = {1,-1,-1,1,1,-1,1,1,1,-1};        int begin,end;        find_range (arr1, 10, begin, end);        //Test 2:        int arr2[10] = {0,0,0,6,0,0,0,0,0,0};        find_range (arr2, 10, begin, end);}