给定一个非负整数数组 nums
和一个整数 m
,你需要将这个数组分成 m
个非空的连续子数组。
设计一个算法使得这 m
个子数组各自和的最大值最小。
示例 1:
输入:nums = [7,2,5,10,8], m = 2 输出:18 解释: 一共有四种方法将 nums 分割为 2 个子数组。 其中最好的方式是将其分为 [7,2,5] 和 [10,8] 。 因为此时这两个子数组各自的和的最大值为18,在所有情况中最小。
示例 2:
输入:nums = [1,2,3,4,5], m = 2 输出:9
示例 3:
输入:nums = [1,4,4], m = 3 输出:4
提示:
1 <= nums.length <= 1000
0 <= nums[i] <= 106
1 <= m <= min(50, nums.length)
二分查找。
二分枚举子数组的和的最大值,找到满足条件的最小值。
class Solution:
def splitArray(self, nums: List[int], m: int) -> int:
def check(x):
s, cnt = 0, 1
for num in nums:
if s + num > x:
cnt += 1
s = num
else:
s += num
return cnt <= m
left, right = max(nums), sum(nums)
while left < right:
mid = (left + right) >> 1
if check(mid):
right = mid
else:
left = mid + 1
return left
class Solution {
public int splitArray(int[] nums, int m) {
int mx = -1;
for (int num : nums) {
mx = Math.max(mx, num);
}
int left = mx, right = (int) 1e9;
while (left < right) {
int mid = (left + right) >> 1;
if (check(nums, m, mid)) {
right = mid;
} else {
left = mid + 1;
}
}
return left;
}
private boolean check(int[] nums, int m, int x) {
int s = 0, cnt = 1;
for (int num : nums) {
if (s + num > x) {
++cnt;
s = num;
} else {
s += num;
}
}
return cnt <= m;
}
}
class Solution {
public:
int splitArray(vector<int>& nums, int m) {
int left = *max_element(nums.begin(), nums.end()), right = (int) 1e9;
while (left < right) {
int mid = left + right >> 1;
if (check(nums, m, mid)) right = mid;
else left = mid + 1;
}
return left;
}
bool check(vector<int>& nums, int m, int x) {
int s = 0, cnt = 1;
for (int num : nums) {
if (s + num > x) {
++cnt;
s = num;
} else {
s += num;
}
}
return cnt <= m;
}
};
func splitArray(nums []int, m int) int {
mx := -1
for _, num := range nums {
mx = max(mx, num)
}
left, right := mx, int(1e9)
for left < right {
mid := (left + right) >> 1
if check(nums, m, mid) {
right = mid
} else {
left = mid + 1
}
}
return left
}
func check(nums []int, m, x int) bool {
s, cnt := 0, 1
for _, num := range nums {
if s+num > x {
cnt++
s = num
} else {
s += num
}
}
return cnt <= m
}
func max(a, b int) int {
if a > b {
return a
}
return b
}