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LeetCode 1013. Partition Array Into Three Parts With Equal Sum (Java版; Easy)

题目描述

Given an array A of integers, return true if and only if we can partition the array into three non-empty parts with equal sums.

Formally, we can partition the array if we can find indexes i+1 < j with 
(A[0] + A[1] + ... + A[i] == A[i+1] + A[i+2] + ... + A[j-1] == A[j] + A[j-1] + ... + A[A.length - 1])

 

Example 1:

Input: A = [0,2,1,-6,6,-7,9,1,2,0,1]
Output: true
Explanation: 0 + 2 + 1 = -6 + 6 - 7 + 9 + 1 = 2 + 0 + 1
Example 2:

Input: A = [0,2,1,-6,6,7,9,-1,2,0,1]
Output: false
Example 3:

Input: A = [3,3,6,5,-2,2,5,1,-9,4]
Output: true
Explanation: 3 + 3 = 6 = 5 - 2 + 2 + 5 + 1 - 9 + 4
 

Constraints:

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

第一次做; 动动脑子, 分析一下题目; 核心: 1)每一个部分的和都得等于sum/3

/*
动动脑子分析一下题目
a=b=c
a+b+c=sum
a=b=c=sum/3
*/
class Solution {
    public boolean canThreePartsEqualSum(int[] A) {
        if(A==null || A.length<3)
            return false;
        int n = A.length;
        int sum = 0;
        for(int a : A)
            sum += a;
        //寻找第一部分
        int cur=0;
        int i=0;
        for(; i<=n-3; i++){
            cur += A[i];
            if(cur==sum/3)
                break;
        }
        //找不到第一部分, 返回false
        if(cur!=sum/3)
            return false;
        
        //寻找第二部分
        cur=0;
        i++;
        for(; i<=n-2; i++){
            cur += A[i];
            if(cur==sum/3)
                break;
        }
        //找不到第二部分, 返回false
        if(cur!=sum/3)
            return false;
        
        
        //是否有第三部分
        return sum - cur*2 == sum/3;
    }
    
}

第一次做; 暴力; 超时50/55; 使用HashMap存储第一部分的和也会超时

class Solution {
    public boolean canThreePartsEqualSum(int[] A) {
        if(A==null || A.length<3)
            return false;
        int n = A.length;
        int sum = 0;
        for(int a : A)
            sum += a;
        // HashMap<Integer, Integer> map = new HashMap<>();
        //i是第一部分的结尾; j是第二部分的结尾
        for(int i=0; i<=n-3; i++){
            int sum1 = getSum(A, 0, i);
            for(int j=i+1; j<=n-2; j++){
                int sum2 = getSum(A, i+1, j);
                int sum3 = sum - sum1 - sum2;
                if(sum1==sum2 && sum1==sum3)
                    return true;
            }
        }
        return false;
    }
    
    private int getSum(int[] arr, int i, int j){
        int sum = 0;
        for(int index=i; index<=j; index++){
            sum += arr[index];
        }
        return sum;
    }
}