添加 problem 719

Algorithms/0719. Find K-th Smallest Pair Distance/719. Find K-th Smallest Pair Distance.go

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+package leetcode
+
+import (
+	"sort"
+)
+
+func smallestDistancePair(nums []int, k int) int {
+	sort.Ints(nums)
+	low, high := 0, nums[len(nums)-1]-nums[0]
+	for low < high {
+		mid := low + (high-low)>>1
+		tmp := findDistanceCount(nums, mid)
+		if tmp >= k {
+			high = mid
+		} else {
+			low = mid + 1
+		}
+	}
+	return low
+}
+
+// 解法一 双指针
+func findDistanceCount(nums []int, num int) int {
+	count, i := 0, 0
+	for j := 1; j < len(nums); j++ {
+		for nums[j]-nums[i] > num && i < j {
+			i++
+		}
+		count += (j - i)
+	}
+	return count
+}
+
+// 解法二 暴力查找
+func findDistanceCount1(nums []int, num int) int {
+	count := 0
+	for i := 0; i < len(nums); i++ {
+		for j := i + 1; j < len(nums); j++ {
+			if nums[j]-nums[i] <= num {
+				count++
+			}
+		}
+	}
+	return count
+}

Algorithms/0719. Find K-th Smallest Pair Distance/719. Find K-th Smallest Pair Distance_test.go

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+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question719 struct {
+	para719
+	ans719
+}
+
+// para 是参数
+// one 代表第一个参数
+type para719 struct {
+	num []int
+	k   int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans719 struct {
+	one int
+}
+
+func Test_Problem719(t *testing.T) {
+
+	qs := []question719{
+
+		question719{
+			para719{[]int{1, 3, 1}, 1},
+			ans719{0},
+		},
+
+		question719{
+			para719{[]int{1, 1, 1}, 2},
+			ans719{0},
+		},
+
+		question719{
+			para719{[]int{1, 6, 1}, 3},
+			ans719{5},
+		},
+
+		question719{
+			para719{[]int{62, 100, 4}, 2},
+			ans719{58},
+		},
+
+		question719{
+			para719{[]int{9, 10, 7, 10, 6, 1, 5, 4, 9, 8}, 18},
+			ans719{2},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 719------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans719, q.para719
+		fmt.Printf("【input】:%v       【output】:%v\n", p, smallestDistancePair(p.num, p.k))
+	}
+	fmt.Printf("\n\n\n")
+}

Algorithms/0719. Find K-th Smallest Pair Distance/README.md

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+# [719. Find K-th Smallest Pair Distance](https://leetcode.com/problems/find-k-th-smallest-pair-distance/)
+
+
+## 题目:
+
+Given an integer array, return the k-th smallest **distance** among all the pairs. The distance of a pair (A, B) is defined as the absolute difference between A and B.
+
+**Example 1:**
+
+    Input:
+    nums = [1,3,1]
+    k = 1
+    Output: 0 
+    Explanation:
+    Here are all the pairs:
+    (1,3) -> 2
+    (1,1) -> 0
+    (3,1) -> 2
+    Then the 1st smallest distance pair is (1,1), and its distance is 0.
+
+**Note:**
+
+1. `2 <= len(nums) <= 10000`.
+2. `0 <= nums[i] < 1000000`.
+3. `1 <= k <= len(nums) * (len(nums) - 1) / 2`.
+
+
+## 题目大意
+
+给定一个整数数组，返回所有数对之间的第 k 个最小距离。一对 (A, B) 的距离被定义为 A 和 B 之间的绝对差值。
+
+提示:
+
+1. 2 <= len(nums) <= 10000.
+2. 0 <= nums[i] < 1000000.
+3. 1 <= k <= len(nums) * (len(nums) - 1) / 2.
+
+
+
+## 解题思路
+
+- 给出一个数组，要求找出第 k 小两两元素之差的值。两两元素之差可能重复，重复的元素之差算多个，不去重。
+- 这一题可以用二分搜索来解答。先把原数组排序，那么最大的差值就是 `nums[len(nums)-1] - nums[0]` ，最小的差值是 0，即在 `[0, nums[len(nums)-1] - nums[0]]` 区间内搜索最终答案。针对每个 `mid`，判断小于等于 `mid` 的差值有多少个。题意就转化为，在数组中找到这样一个数，使得满足 `nums[i] - nums[j] ≤ mid` 条件的组合数等于 `k`。那么如何计算满足两两数的差值小于 mid 的组合总数是本题的关键。
+- 最暴力的方法就是 2 重循环，暴力计数。这个方法效率不高，耗时很长。原因是没有利用数组有序这一条件。实际上数组有序对计算满足条件的组合数有帮助。利用双指针滑动即可计算出组合总数。见解法一。