410. Split Array Largest Sum

最新推荐文章于 2024-10-24 16:28:26 发布

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最新推荐文章于 2024-10-24 16:28:26 发布

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本文介绍了一种将数组划分为m个连续子数组，并使这些子数组中最大和最小化的算法实现。通过对最终结果进行二分查找，该算法有效地解决了问题。

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Given an array which consists of non-negative integers and an integer m, you can split the array into m non-empty continuous subarrays. Write an algorithm to minimize the largest sum among these m subarrays.

Note:
If n is the length of array, assume the following constraints are satisfied:

1 ≤ n ≤ 1000
1 ≤ m ≤ min(50, n)

Examples:

Input:
nums = [7,2,5,10,8]
m = 2

Output:
18

Explanation:
There are four ways to split nums into two subarrays.
The best way is to split it into [7,2,5] and [10,8],
where the largest sum among the two subarrays is only 18.

思路：这种求最大值最小化的问题，就是BS的讨论，对最后可能的结果直接进行二分，然后判断是否可以到达mid，如此而已

public class Solution {
    public int splitArray(int[] nums, int m) {
        int max = Integer.MIN_VALUE, sum = 0;
        for(int i : nums)	{
        	sum += i;
        	max = Math.max(max, i);
        }
        
        // BS
        int lo = max, hi = sum;
        while(lo <= hi) {
        	int mid = lo + (hi - lo) / 2;
        	if(canGet(nums, mid, m))	lo = mid + 1;
        	else						hi = mid - 1;
        }
        
        return lo;
    }

	private boolean canGet(int[] nums, int mid, int m) {
		int cnt = 0, sum = 0;
		for(int i=0; i<nums.length; i++) {
			if(sum + nums[i] > mid) {
				cnt ++;
				if(cnt == m)	return true;
				sum = nums[i];
				continue;
			}
			
			sum += nums[i];
		}
		
		return false;
	}
}