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410. 分割数组的最大值

程序员文章站 2022-07-01 17:17:39
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410. 分割数组的最大值

题解:

410. 分割数组的最大值

410. 分割数组的最大值

链接:https://leetcode-cn.com/problems/split-array-largest-sum/

下面这种情况是dfs是超时的,有时间加一下记忆化

class Solution {
public:
    int splitArray(vector<int>& nums, int m) {
        int n = nums.size();
        vector<vector<long long>> f(n + 1, vector<long long>(m + 1, LLONG_MAX));
        vector<long long> sub(n + 1, 0);
        for (int i = 0; i < n; i++) {
            sub[i + 1] = sub[i] + nums[i];
        }
        f[0][0] = 0;
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= min(i, m); j++) {
                for (int k = 0; k < i; k++) {
                    f[i][j] = min(f[i][j], max(f[k][j - 1], sub[i] - sub[k]));
                }
            }
        }
        return (int)f[n][m];
    }
};
class Solution {
public:
    int splitArray(vector<int>& nums, int m) {
        if(nums.size() <= 0) {
            return 0;
        }
        int begin = 0;
        int step = m;
        int min_sum = INT_MAX;
        // 后缀数组和
        vector<long> prefix_sum(nums.size(), 0);
        prefix_sum[nums.size()-1] = nums[nums.size()-1];
        if(nums.size() >= 2) {
            for(int i = nums.size()-2; i >= 0; --i) {
                if(prefix_sum[i] == INT_MAX || nums[i] == INT_MAX) {
                    prefix_sum[i] = INT_MAX;
                } else {
                    prefix_sum[i] = nums[i] + prefix_sum[i+1];
                }
            }
        }
        vector<int> path;
        dfs(nums, path, begin, step, min_sum, prefix_sum);
        return min_sum;
    }
    void dfs(vector<int>& nums, vector<int>& path, int begin, int step, int& min_sum
            , vector<long>& prefix_sum) {
        if(begin == nums.size()) {
            return;
        }
        if(step == 1) {
            path.push_back(prefix_sum[begin]);
            // 取出这种分割情况下的最大值
            auto max_ite = max_element(path.begin(), path.end());
            int max_value = *max_ite;
            // 判断最小值
            min_sum = min(min_sum, max_value);
            path.pop_back();
            return;
        }
        int temp_sum = 0;
        for(int i = begin; i < nums.size(); ++i) {
            if(temp_sum == INT_MAX || nums[i] == INT_MAX) {
                temp_sum = INT_MAX;
            } else {
                temp_sum += nums[i];
            }
            path.push_back(temp_sum);
            dfs(nums, path, i+1, step-1, min_sum, prefix_sum);
            path.pop_back();
        }
    }
};

 

相关标签: leetcode