2019-10-14 22:13:18
问题描述:
问题求解:
经典的动态规划问题。
public int splitArray(int[] nums, int m) { int n = nums.length; long[][] dp = new long[m + 1][n]; long[] presum = new long[n]; long curSum = 0; for (int i = 0; i < n; i++) { curSum += nums[i]; presum[i] = curSum; dp[1][i] = curSum; } for (int i = 2; i <= m; i++) { for (int j = 0; j < n; j++) { dp[i][j] = curSum; for (int k = 0; k < j; k++) { dp[i][j] = Math.min(dp[i][j], Math.max(dp[i - 1][k], presum[j] - presum[k])); } } } return (int)dp[m][n - 1]; }
动态规划-划分数组的最大和 Split Array Largest Sum
原文:https://www.cnblogs.com/hyserendipity/p/11674576.html