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.

http://www.cnblogs.com/grandyang/p/5933787.html
https://leetcode.com/problems/split-array-largest-sum/discuss/89817

Solution：

思路：
Use binary search to approach the correct answer. We have l = max number of array; r = sum of all numbers in the array;Every time we do mid = (l + r) / 2;
接下来要做的是找出 和最大 且 小于等于mid的子数组的个数
每组最大为target，
如果我们无法分为m组，说明target偏大，再二分调整；
反之，说明m偏小，再二分调整

Time Complexity: O() Space Complexity: O()

Solution Code：

public class Solution {
    public int splitArray(int[] nums, int m) {
        int max = 0; long sum = 0;
        for (int num : nums) {
            max = Math.max(num, max);
            sum += num;
        }
        if (m == 1) return (int)sum;
        //binary search
        long l = max; long r = sum;
        while (l <= r) {
            long mid = (l + r)/ 2;
            if (valid(mid, nums, m)) {
                r = mid - 1;
            } else {
                l = mid + 1;
            }
        }
        return (int)l;
    }
    public boolean valid(long target, int[] nums, int m) {
        int count = 1;
        long total = 0;
        for(int num : nums) {
            total += num;
            if (total > target) {
                total = num;
                count++;
                if (count > m) {
                    return false;
                }
            }
        }
        return true;
    }
}