LeetCode No.410 Split Array Largest Sum - 黄健伟的博客 - CSDN博客
LeetCode No.410 Split Array Largest Sum
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.
Given&m&satisfies the following constraint: 1 ≤ m ≤ length(nums) ≤ 14,000.
nums = [7,2,5,10,8]
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.
====================================================================================================================================
此题目的大意是：将给定数组切割成m份，求这m份中最大和的最小值（即求极大的极小问题）。
因为我们知道切割的方法，用二分+搜索求解就可以了。
所谓二分，就是确定结果的最大值maxerf和最小值miner，取两者的中间值mid。而搜索便是：判断mid是否符合条件，如果符合，则令miner=mid+1，同时记录下ans，否则，令maxer=mid-1，以此类推，当miner&maxer时结束。
方法不难，但是我们还要注意一些细节上的问题：
（1）nums = [1,] ，m = 1 时，计算maxer的时候会越界，而且计算繁杂，应单独处理
（2）nums = [1,] ，m = 2 时，计算maxer的时候会越界，得到的maxer为负数，这时候，要令maxer =&。
（3）在搜索的时候，当当前和temp==mid时，要判断此时是否为数组的结尾，再确定sum是否加1。
class Solution {
int splitArray(vector&int&& nums, int m) {
int n = nums.size() ;
if ( n == 0 )
return 0 ;
int miner = 0 , maxer = 0 ;
for ( int i = 0 ; i & i ++ )
maxer += nums[i] ;
miner = max ( miner , nums[i] ) ;
if ( m == 1 )
if ( maxer & 0 )
miner = max ( miner , maxer / m ) ;
while ( miner &= maxer )
int mid = ( (unsigned int) miner + maxer ) / 2 , temp = 0 , sum = 1 ;
for ( int i = 0 ; i & i ++ )
temp += nums[i] ;
if ( temp & mid )
temp = nums[i] ;
else if ( temp == mid )
if ( i != n - 1 )
temp = 0 ;
if ( sum &= m )
maxer = mid - 1 ;
miner = mid + 1 ;
我的热门文章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.
Given&m&satisfies the following constraint: 1 ≤ m ≤ length(nums) ≤ 14,000.
nums = [7,2,5,10,8]
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.
dp[i][j]表示nums数组从index=0到index=i，切j刀最小的和，dp[i][j]=min(max(dp[i][k],sums[i]-sums[k]))，其中j-1&=k&=i-1，
然而二维数组开不下。
public class Solution {
public static int splitArray(int[] nums, int m)
int len=nums.
int[] sums=new int[len];
int sum=0;
for(int i=0;i&i++)
sum+=nums[i];
int[][] dp=new int[len][len];
for(int i=0;i&i++)
dp[i][0]=sums[i];
for(int i=0;i&i++)
for(int j=1;j&j++)
int min=Integer.MAX_VALUE;
for(int k=i-1;k&=j-1;k--)
min=Math.min(min, Math.max(dp[k][j-1], sums[i]-sums[k]));
return dp[len-1][m-1];
----------------------------------------------------------------------
二分搜索法，摘自
/topic/61324/clear-explanation-8ms-binary-search-java/9
The answer is between maximum value of input array numbers and sum of those numbers.Use binary search to approach the correct answer. We have&l = r
= sum of all Every time we do&mid = (l + r) / 2;Use greedy to narrow down left and right boundaries in binary search.
3.1 Cut the array from left.
3.2 Try our best to make sure that the sum of numbers between each two cuts (inclusive) is large enough but still less thanmid.
3.3 We'll end up with two results: either we can divide the array into more than m subarrays or we cannot.
If we can, it means that the&mid&value we pick is too small because we've already tried our best to make sure each
part holds as many non-negative numbers as we can but we still have numbers left. So, it is impossible to cut the array into m parts and make sure each parts is no larger than&mid.
We should increase m. This leads to&l = mid + 1;
If we can't, it is either we successfully divide the array into m parts and the sum of each part is less than&mid,
or we used up all numbers before we reach m. Both of them mean that we should lower&mid&because we need to find the minimum one. This
leads to&r = mid - 1;
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 +=
if (m == 1) return (int)
long l = long r =
while (l &= r) {
long mid = (l + r)/ 2;
if (valid(mid, nums, m)) {
r = mid - 1;
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 +=
if (total & target) {
count++;
if (count & m) {
return false;
return true;
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