LeetCode-1013. 将数组分成和相等的三个部分

给你一个整数数组 A,只有可以将其划分为三个和相等的非空部分时才返回 true,否则返回 false。

形式上,如果可以找出索引 i+1 < j 且满足 (A[0] + A[1] + ... + A[i] == A[i+1] + A[i+2] + ... + A[j-1] == A[j] + A[j-1] + ... + A[A.length - 1]) 就可以将数组三等分。

 

示例 1:

输出:[0,2,1,-6,6,-7,9,1,2,0,1]
输出:true
解释:0 + 2 + 1 = -6 + 6 - 7 + 9 + 1 = 2 + 0 + 1
示例 2:

输入:[0,2,1,-6,6,7,9,-1,2,0,1]
输出:false
示例 3:

输入:[3,3,6,5,-2,2,5,1,-9,4]
输出:true
解释:3 + 3 = 6 = 5 - 2 + 2 + 5 + 1 - 9 + 4
 

提示:

3 <= A.length <= 50000
-10^4 <= A[i] <= 10^4

 

每日打卡:

#include 
#include 
using namespace std;
class Solution {
public:
	bool canThreePartsEqualSum(vector& A) {
		int flag1 = false;
		int flag2 = false;
		int flag3 = false;

		int aver = 0;
		for (int i = 0; i < A.size(); i++) {
			aver += A[i];
		}

		if (aver % 3 != 0)
			return false;

		aver /= 3;

		int i = 0;
		int sum = 0;
		for (i = 0; i < A.size(); i++) {
			sum += A[i];
			if (sum == aver) {
				flag1 = true;
				break;
			}
		}

		int j = 0;
		sum = 0;
		for (j = i + 1; j < A.size(); j++) {
			sum += A[j];
			if (sum == aver) {
				flag2 = true;
				break;
			}
		}


		int k = 0;
		sum = 0;
		for (k = j + 1; k < A.size(); k++) {
			sum += A[k];
		}

		if (sum == aver && (j != A.size() - 1)) {
			flag3 = true;
		}


		return (flag1 && flag2 && flag3);
	}

private:
	void quicksort(vector& a, int left, int right) {
		int i = left;
		int j = right;
		int key = a[i];
		if (left < right) {
			while (i < j) {
				while (i < j && a[j] >= key) {
					j--;
				}
				a[i] = a[j];
				while (i < j && a[i] <= key) {
					i++;
				}
				a[j] = a[i];
			}
			a[i] = key;
			quicksort(a, left, j - 1);
			quicksort(a, i + 1, right);
		}
		else {
			return;
		}
	}
};



int main() {
	vectora1 = { 0,2,1,-6,6,-7,9,1,2,0,1 };
	vectora2 = { 1, -1, 1, -1 };

	Solution* ps = new Solution();
	cout << ps->canThreePartsEqualSum(a2) << endl;

	return 0;
}

 

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