codility problem - NumberOfDiscIntersections

就是求 区间覆盖的问题。

【x， y】

按x排序， 对y，二分找刚好大于它的x。

package solution

// you can also use imports, for example:
import "fmt"
import "sort"

// you can write to stdout for debugging purposes, e.g.
// fmt.Println("this is a debug message")

type interval struct {
    x, y int64
}

func (i interval) String() string {
    return fmt.Sprintf("[%d, %d]", i.x, i.y)
}

type ByX []interval

func (a ByX) Len() int           { return len(a) }
func (a ByX) Swap(i, j int)      { a[i], a[j] = a[j], a[i] }
func (a ByX) Less(i, j int) bool { return a[i].x < a[j].x }

func Solution(A []int) int {
    N := len(A)
    intervals := []interval{}
    for i, a := range(A) {
        intervals = append(intervals, interval{int64(i - a), int64(i + a)})
    }
    sort.Sort(ByX(intervals))
    //fmt.Println(intervals)
    count := 0
    for i := 0; i < N - 1; i++ {
        target := intervals[i].y
        low, high := i + 1, N
        for ; low < high; {
            mid := low + ((high - low) >> 1)
            if intervals[mid].x <= target {
                low = mid + 1
            } else {
                high = mid
            }
        }
        //fmt.Println("i: ", i, "##", intervals[i])
        //fmt.Println("low: ", low)
        count += low - i - 1
        //fmt.Println("count:", count)
    }
    return count
}

原文地址：https://www.cnblogs.com/brayden/p/9027066.html