Add solution 0119、1439

leetcode/0119.Pascals-Triangle-II/119. Pascal's Triangle II.go

@@ -0,0 +1,10 @@
+package leetcode
+
+func getRow(rowIndex int) []int {
+	row := make([]int, rowIndex+1)
+	row[0] = 1
+	for i := 1; i <= rowIndex; i++ {
+		row[i] = row[i-1] * (rowIndex - i + 1) / i
+	}
+	return row
+}

leetcode/0119.Pascals-Triangle-II/119. Pascal's Triangle II_test.go

@@ -0,0 +1,47 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question119 struct {
+	para119
+	ans119
+}
+
+// para 是参数
+// one 代表第一个参数
+type para119 struct {
+	rowIndex int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans119 struct {
+	one []int
+}
+
+func Test_Problem119(t *testing.T) {
+
+	qs := []question119{
+
+		{
+			para119{3},
+			ans119{[]int{1, 3, 3, 1}},
+		},
+
+		{
+			para119{0},
+			ans119{[]int{1}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 119------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans119, q.para119
+		fmt.Printf("【input】:%v    【output】:%v\n", p, getRow(p.rowIndex))
+	}
+	fmt.Printf("\n\n\n")
+}

leetcode/0119.Pascals-Triangle-II/README.md

@@ -0,0 +1,72 @@
+# [119. Pascal's Triangle II](https://leetcode.com/problems/pascals-triangle-ii/)
+
+
+## 题目
+
+Given an integer `rowIndex`, return the `rowIndexth` row of the Pascal's triangle.
+
+Notice that the row index starts from **0**.
+
+![https://upload.wikimedia.org/wikipedia/commons/0/0d/PascalTriangleAnimated2.gif](https://upload.wikimedia.org/wikipedia/commons/0/0d/PascalTriangleAnimated2.gif)
+
+In Pascal's triangle, each number is the sum of the two numbers directly above it.
+
+**Follow up:**
+
+Could you optimize your algorithm to use only *O*(*k*) extra space?
+
+**Example 1:**
+
+```
+Input: rowIndex = 3
+Output: [1,3,3,1]
+```
+
+**Example 2:**
+
+```
+Input: rowIndex = 0
+Output: [1]
+```
+
+**Example 3:**
+
+```
+Input: rowIndex = 1
+Output: [1,1]
+```
+
+**Constraints:**
+
+- `0 <= rowIndex <= 33`
+
+## 题目大意
+
+给定一个非负索引 k，其中 k ≤ 33，返回杨辉三角的第 k 行。
+
+## 解题思路
+
+- 题目中的三角是杨辉三角，每个数字是 `(a+b)^n` 二项式展开的系数。题目要求我们只能使用 O(k) 的空间。那么需要找到两两项直接的递推关系。由组合知识得知：
+
+    $$\begin{aligned}C_{n}^{m} &= \frac{n!}{m!(n-m)!} \\C_{n}^{m-1} &= \frac{n!}{(m-1)!(n-m+1)!}\end{aligned}$$
+
+    于是得到递推公式：
+
+    $$C_{n}^{m} = C_{n}^{m-1} \times \frac{n-m+1}{m}$$
+
+    利用这个递推公式即可以把空间复杂度优化到 O(k)
+
+## 代码
+
+```go
+package leetcode
+
+func getRow(rowIndex int) []int {
+	row := make([]int, rowIndex+1)
+	row[0] = 1
+	for i := 1; i <= rowIndex; i++ {
+		row[i] = row[i-1] * (rowIndex - i + 1) / i
+	}
+	return row
+}
+```

leetcode/1439.Find-the-Kth-Smallest-Sum-of-a-Matrix-With-Sorted-Rows/1439. Find the Kth Smallest Sum of a Matrix With Sorted Rows.go

@@ -0,0 +1,82 @@
+package leetcode
+
+import "container/heap"
+
+func kthSmallest(mat [][]int, k int) int {
+	if len(mat) == 0 || len(mat[0]) == 0 || k == 0 {
+		return 0
+	}
+	prev := mat[0]
+	for i := 1; i < len(mat); i++ {
+		prev = kSmallestPairs(prev, mat[i], k)
+	}
+	if k < len(prev) {
+		return -1
+	}
+	return prev[k-1]
+}
+
+func kSmallestPairs(nums1 []int, nums2 []int, k int) []int {
+	res := []int{}
+	if len(nums2) == 0 {
+		return res
+	}
+	pq := newPriorityQueue()
+	for i := 0; i < len(nums1) && i < k; i++ {
+		heap.Push(pq, &pddata{
+			n1:    nums1[i],
+			n2:    nums2[0],
+			n2Idx: 0,
+		})
+	}
+	for pq.Len() > 0 {
+		i := heap.Pop(pq)
+		data := i.(*pddata)
+		res = append(res, data.n1+data.n2)
+		k--
+		if k <= 0 {
+			break
+		}
+		idx := data.n2Idx
+		idx++
+		if idx >= len(nums2) {
+			continue
+		}
+		heap.Push(pq, &pddata{
+			n1:    data.n1,
+			n2:    nums2[idx],
+			n2Idx: idx,
+		})
+	}
+	return res
+}
+
+type pddata struct {
+	n1    int
+	n2    int
+	n2Idx int
+}
+
+type priorityQueue []*pddata
+
+func newPriorityQueue() *priorityQueue {
+	pq := priorityQueue([]*pddata{})
+	heap.Init(&pq)
+	return &pq
+}
+
+func (pq priorityQueue) Len() int           { return len(pq) }
+func (pq priorityQueue) Swap(i, j int)      { pq[i], pq[j] = pq[j], pq[i] }
+func (pq priorityQueue) Less(i, j int) bool { return pq[i].n1+pq[i].n2 < pq[j].n1+pq[j].n2 }
+func (pq *priorityQueue) Pop() interface{} {
+	old := *pq
+	val := old[len(old)-1]
+	old[len(old)-1] = nil
+	*pq = old[0 : len(old)-1]
+	return val
+}
+
+func (pq *priorityQueue) Push(i interface{}) {
+	val := i.(*pddata)
+	*pq = append(*pq, val)
+}

leetcode/1439.Find-the-Kth-Smallest-Sum-of-a-Matrix-With-Sorted-Rows/1439. Find the Kth Smallest Sum of a Matrix With Sorted Rows_test.go

@@ -0,0 +1,56 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1439 struct {
+	para1439
+	ans1439
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1439 struct {
+	mat [][]int
+	k   int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1439 struct {
+	one int
+}
+
+func Test_Problem1439(t *testing.T) {
+
+	qs := []question1439{
+
+		{
+			para1439{[][]int{{1, 3, 11}, {2, 4, 6}}, 5},
+			ans1439{7},
+		},
+
+		{
+			para1439{[][]int{{1, 3, 11}, {2, 4, 6}}, 9},
+			ans1439{17},
+		},
+		{
+			para1439{[][]int{{1, 10, 10}, {1, 4, 5}, {2, 3, 6}}, 7},
+			ans1439{9},
+		},
+		{
+			para1439{[][]int{{1, 1, 10}, {2, 2, 9}}, 7},
+			ans1439{12},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1439------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1439, q.para1439
+		fmt.Printf("【input】:%v       【output】:%v\n", p, kthSmallest(p.mat, p.k))
+	}
+	fmt.Printf("\n\n\n")
+}