[LintCode] Maximum Subarray III

最新推荐文章于 2019-01-21 16:26:34 发布

转载最新推荐文章于 2019-01-21 16:26:34 发布 · 118 阅读

本文介绍了一种寻找数组中k个非重叠子数组的最大和的算法，通过动态规划实现，详细解析了算法思路和代码实现，展示了如何优化空间复杂度。

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Maximum Subarray III

Given an array of integers and a number k, find knon-overlapping subarrays which have the largest sum.

The number in each subarray should be contiguous.

Return the largest sum.

Example

Given [-1,4,-2,3,-2,3], k=2, return 8

Note

The subarray should contain at least one number

Analysis:

DP. d[i][j] means the maximum sum we can get by selecting j subarrays from the first i elements.

d[i][j] = max{d[p][j-1]+maxSubArray(p+1,i)}

we iterate p from i-1 to j-1, so we can record the max subarray we get at current p, this value can be used to calculate the max subarray from p-1 to i when p becomes p-1.

 1 class Solution {
 2 public:
 3     /**
 4      * @param nums: A list of integers
 5      * @param k: An integer denote to find k non-overlapping subarrays
 6      * @return: An integer denote the sum of max k non-overlapping subarrays
 7      */
 8     int maxSubArray(vector<int> nums, int k) {
 9         // write your code here
10         int n = nums.size();
11         vector<vector<int>> dp(n + 1, vector<int>(k + 1, INT_MIN));
12         for (int i = 0; i <= n; ++i) 
13             dp[i][0] = 0;
14         for (int i = 1; i <= n; ++i) {
15             for (int j = 1; j <= min(i, k); ++j) {
16                 int tmp = 0;
17                 for (int t = i - 1; t >= j - 1; --t) {
18                     tmp = max(tmp + nums[t], nums[t]);
19                     dp[i][j] = max(dp[i][j], tmp + dp[t][j-1]);
20                 }
21             }
22         }
23         return dp[nums.size()][k];
24     }
25 };

重复使用dp数组，可以将空间复杂度降到O(n)。为了避免冲突，在枚举求解数组长度n时到倒着求，这样可以保证上一次迭代的结果不被覆盖掉。

 1 class Solution {
 2 public:
 3     /**
 4      * @param nums: A list of integers
 5      * @param k: An integer denote to find k non-overlapping subarrays
 6      * @return: An integer denote the sum of max k non-overlapping subarrays
 7      */
 8     int maxSubArray(vector<int> nums, int k) {
 9         // write your code here
10         int n = nums.size();
11         vector<int> dp(n + 1, 0);
12         for (int j = 1; j <= k; ++j) {
13             for (int i = n; i >= j; --i) {
14                 dp[i] = INT_MIN;
15                 int tmp = 0;
16                 for (int t = i - 1; t >= j - 1; --t) {
17                     tmp = max(tmp + nums[t], nums[t]);
18                     dp[i] = max(dp[i], tmp + dp[t]);
19                 }
20             }
21         }
22         return dp[n];
23     }
24 };