[leetcode/lintcode 题解] 算法面试真题详解：最大线段覆盖

描述
在一个数轴上给出n个线段，问选择不超过k个线段，使得这k个线段覆盖的数最多。

在线评测地址：领扣题库官网

样例1
Input:
[(1,2),(2,3),(3,4)]
2
Output: 4
Explanation:
Select the line segment (1,2), (3,4), which can cover the 4 numbers of 1,2,3,4.

样例2
Input:
[(1,2),(2,3),(1,7)]
2
Output: 7
Explanation:
Selecting the line segment (1,7) ,which can cover the 7 numbers of 1,2,3,4,5,6,7.

题解
dpidpi表示用jj个线段覆盖前ii个数的最优答案。先将所有线段按照左端点排序，对于左端点相同的线段，取最长的拿来转移。 则有： dpi+1=max(dpi,dpi+1)dpi+1=max(dpi,dpi+1) dpi+num=max(dpi+num,dpi+num)dpi+num=max(dpi+num,dpi+num)(num为线段长度)

/**
 * Definition of Interval:
 * public classs Interval {
 *     int start, end;
 *     Interval(int start, int end) {
 *         this.start = start;
 *         this.end = end;
 *     }
 * }
 */

public class Solution {
    /**
     * @param intervals: The intervals
     * @param k: The k
     * @return: The answer
     */
    
    class Cmp implements Comparator<Interval>{
       @Override
       public int compare(Interval a, Interval b){
            if(a.start == b.start) {
                return a.end - b.end;
            }
            return a.start - b.start;
        }
    } 
    
    public int maximumLineCoverage(List<Interval> intervals, int k) {
        // Write your code here
        Collections.sort(intervals, new Cmp());
        int index = 0;
        int num = 0;
        int maxnum = 0;
        int[][] dp = new int[2005][2005];
        for (int i = 0; i < intervals.size(); i++) {
            maxnum = Math.max(intervals.get(i).end, maxnum);
        }
        for (int i = 0; i < maxnum; i++) {
            while (index < intervals.size() && intervals.get(index).start == i + 1) {
                num = Math.max(num, intervals.get(index).end - intervals.get(index).start + 1);
                index++;
            }
            for (int j = 0; j <= k; j++) {
                dp[i + 1][j] = Math.max(dp[i][j], dp[i + 1][j]);//不取
                if (i + num <= maxnum) {
                    dp[i + num][j + 1] = Math.max(dp[i][j] + num, dp[i + num][j + 1]);//取
                }
            }
            if (num > 0) {//i加了1，所以线段长度减1
                num--;
            }
        }
        return dp[maxnum][k];
    }
}

更多题解参考：九章官网solution