1 兩次遍歷 尋找最大和

2 貪心思想，sum為負數，sum為當前值， sum大於0,為sum加當前值; 如果全為負數，找到最大的一個負數

//連續子陣列的最大和
    int FindGreatSum(vector<int> array) {
        if (array.size() == 0) {
            return 0;
        }
        int sum = 0 , maxsum = 0;
        for (int i = 0; i < array.size(); i++) {
            sum = 0;//每次一開始重置0
            for(int j = i; j < array.size(); j++) {

                sum = sum + array[j];
                if (sum > maxsum) {
                    maxsum = sum;
                    cout << maxsum << endl;
                }
            }
        }
        return maxsum;
    }

    //貪心思想
    int FindGreatSum2(vector<int> array) {
        if (array.size() == 0) {
            return 0;
        }

        int sum = 0, maxsum = -100, maxnum = -10000;
        for (int i = 0; i < array.size(); i++) {
            if (sum <= 0) {
                sum = array[i];//和小於0,和就為當前數
            }
            else {
                sum = sum + array[i];//大於0，加上當前數
            }

            if (sum > maxsum) {
                maxsum = sum;
            }
            if (array[i] > maxnum) { // 全負數情況
                maxnum = array[i];
            }
        }
        return (maxnum < 0 ) ? maxnum : maxsum;

    }