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:
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.
分析
这道题如果用深搜的话一定会超时的。所以只能使用动态规划。动态规划怎么确定数组的含义就比较重要了。
我们使用dp[i][j]表示nums[0:j]的数组被分割i 次后所得到的最小的sum。那么可以推导出:
dp[i][j] = min(max(dp[i-1][k], sum[k:m])), i <= k < j。
即我们计算i次分割的结果时,我们先去找i-1次分割的结果,然后计算i-1次分割的结果与sum[k:j]的最大值,最后遍历 i <= k < j时的最小值。
Code
class Solution {
public:
int splitArray(vector& nums, int m) {
int len = nums.size();
if (len == 0)
return 0;
vector> dp(m, vector(len, INT_MAX));
dp[0][0] = nums[0];
for (int i = 1; i < len; i ++)
{
dp[0][i] = dp[0][i-1] + nums[i];
}
for (int i = 1; i < m; i ++)
{
for (int j = i; j < len; j ++)
{
for (int k = i-1; k < j; k ++)
{
dp[i][j] = min(dp[i][j], max(dp[i-1][k], dp[0][j] - dp[0][k]));
}
}
}
return dp[m-1][len-1];
}
};
运行效率
Runtime: 132 ms, faster than 27.65% of C++ online submissions for Split Array Largest Sum.
Memory Usage: 9.2 MB, less than 28.31% of C++ online submissions for Split Array Largest Sum.