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# 915. Partition Array into Disjoint Intervals

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Given an array `nums`, partition it into two (contiguous) subarrays `left` and `right` so that:

• Every element in `left` is less than or equal to every element in `right`.
• `left` and `right` are non-empty.
• `left` has the smallest possible size.

Return the length of `left` after such a partitioning.  It is guaranteed that such a partitioning exists.

Example 1:

```Input: nums = [5,0,3,8,6]
Output: 3
Explanation: left = [5,0,3], right = [8,6]
```

Example 2:

```Input: nums = [1,1,1,0,6,12]
Output: 4
Explanation: left = [1,1,1,0], right = [6,12]
```

Note:

1. `2 <= nums.length <= 30000`
2. `0 <= nums[i] <= 106`
3. It is guaranteed there is at least one way to partition `nums` as described.
```   public int partitionDisjoint(int[] A) {
int maxUntilI = A;
int leftPartitionMax=A;
int partitionSpot = 0;
for(int i=1;i<A.length;i++){
maxUntilI =Math.max(maxUntilI, A[i]) ;
if(A[i]<leftPartitionMax){
leftPartitionMax=maxUntilI;
partitionSpot =i;
}
}
return partitionSpot+1;
} ```

https://leetcode.com/problems/partition-array-into-disjoint-intervals/discuss/175945/Java-one-pass-7-lines