Create 1013. Partition Array Into Three Parts With Equal Sum

Author: haibarawu

1013. Partition Array Into Three Parts With Equal Sum

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+/******************************
+1013. Partition Array Into Three Parts With Equal Sum
+
+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: [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: [0,2,1,-6,6,7,9,-1,2,0,1]
+Output: false
+
+Example 3:
+Input: [3,3,6,5,-2,2,5,1,-9,4]
+Output: true
+Explanation: 3 + 3 = 6 = 5 - 2 + 2 + 5 + 1 - 9 + 4
+
+Note:
+1. 3 <= A.length <= 50000
+2. -10000 <= A[i] <= 10000
+    
+******************************/
+
+/******************************
+1、先遍历一遍，求出数组和，检查总和是否能被3整除;
+2、循环遍历数组A，计算和的一部分partitionSum;如果找到和 sum/3 相等，则将partitionSum重置为0，并增加计数器count;
+3、到最后，如果平均可以看到至少3次，返回true;否则返回false。
+时间复杂度：O(n)
+空间复杂度：O(1)
+******************************/
+
+
+class Solution {
+    public boolean canThreePartsEqualSum(int[] A) {
+        int sum = 0;
+        
+        for (int num : A) {
+            sum += num;
+        }
+        
+        if (sum % 3 != 0) {
+            return false;
+        }
+        
+        int count = 0;
+        int partitionSum = 0;
+        
+        for (int i = 0; i < A.length; i ++) {
+            partitionSum += A[i];
+            
+            if (partitionSum == sum / 3) {
+                partitionSum = 0;
+                count ++;
+            }
+            
+            if (count >= 3) {
+                return true; 
+            }
+        }
+        
+        return false;
+    }
+}