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