剑指offer-面试题31 求连续子数组的最大和

#include<iostream>using namespace std;//常规方法int FindMaxSumOfSubArray1(int Array[], int length){    if(Array == NULL || length <= 0)    {        throw exception("ERROR");    }    int nMax = 0;    int nCurSum = 0;    for(int i = 0; i< length; ++i)    {        if(nCurSum<=0)            nCurSum = Array[i];        else            nCurSum +=Array[i];        if(nCurSum > nMax)        {            nMax = nCurSum;        }    }    return nMax;}int FindMaxSumOfSubArray2(int Array[], int length){    if(Array == NULL || length <= 0)    {        throw exception("ERROR");    }    //运用动态规划,f(i)表示以array[i]结尾的子数组的最大和, 求出所有的f(i)的最大值就行    //如果f(i-1)<=0, 则f(i)=array[i], 否则f(i)=f(i-1)+array[i];    int nPre = Array[0];   //表示以前一个数结尾的序列的最大值    int nCur = 0;   //表示以当前数结尾的序列的最大值    int nMax = 0x80000000;//代表int型最小的负数, 用于保存f(i)的最大值    for(int i = 1; i< length; ++i)    {        if(nPre <= 0)            nCur = Array[i];        else            nCur = nPre+Array[i];        nPre = nCur;        if(nCur > nMax)            nMax = nCur;    }    return nMax;}int main(){    int a[] = {1, -2, 3, 10, -4, 7, 2, -5};    cout<<FindMaxSumOfSubArray2(a, 8)<<endl;}