1. Add solution 1091、1614、1619、1624、1629、1636、1704

2. ctl strings.TrimSpace question.Title

leetcode/1091.Shortest-Path-in-Binary-Matrix/1091. Shortest Path in Binary Matrix.go

@@ -0,0 +1,54 @@
+package leetcode
+
+var dir = [][]int{
+	{-1, -1},
+	{-1, 0},
+	{-1, 1},
+	{0, 1},
+	{0, -1},
+	{1, -1},
+	{1, 0},
+	{1, 1},
+}
+
+func shortestPathBinaryMatrix(grid [][]int) int {
+	visited := make([][]bool, 0)
+	for range make([]int, len(grid)) {
+		visited = append(visited, make([]bool, len(grid[0])))
+	}
+	dis := make([][]int, 0)
+	for range make([]int, len(grid)) {
+		dis = append(dis, make([]int, len(grid[0])))
+	}
+	if grid[0][0] == 1 {
+		return -1
+	}
+	if len(grid) == 1 && len(grid[0]) == 1 {
+		return 1
+	}
+
+	queue := []int{0}
+	visited[0][0], dis[0][0] = true, 1
+	for len(queue) > 0 {
+		cur := queue[0]
+		queue = queue[1:]
+		curx, cury := cur/len(grid[0]), cur%len(grid[0])
+		for d := 0; d < 8; d++ {
+			nextx := curx + dir[d][0]
+			nexty := cury + dir[d][1]
+			if isInBoard(grid, nextx, nexty) && !visited[nextx][nexty] && grid[nextx][nexty] == 0 {
+				queue = append(queue, nextx*len(grid[0])+nexty)
+				visited[nextx][nexty] = true
+				dis[nextx][nexty] = dis[curx][cury] + 1
+				if nextx == len(grid)-1 && nexty == len(grid[0])-1 {
+					return dis[nextx][nexty]
+				}
+			}
+		}
+	}
+	return -1
+}
+
+func isInBoard(board [][]int, x, y int) bool {
+	return x >= 0 && x < len(board) && y >= 0 && y < len(board[0])
+}

leetcode/1091.Shortest-Path-in-Binary-Matrix/1091. Shortest Path in Binary Matrix_test.go

@@ -0,0 +1,48 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1091 struct {
+	para1091
+	ans1091
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1091 struct {
+	grid [][]int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1091 struct {
+	one int
+}
+
+func Test_Problem1091(t *testing.T) {
+
+	qs := []question1091{
+
+		{
+			para1091{[][]int{{0, 1}, {1, 0}}},
+			ans1091{2},
+		},
+
+		{
+			para1091{[][]int{{0, 0, 0}, {1, 1, 0}, {1, 1, 0}}},
+			ans1091{4},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1091------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1091, q.para1091
+		fmt.Printf("【input】:%v       【output】:%v\n", p, shortestPathBinaryMatrix(p.grid))
+
+	}
+	fmt.Printf("\n\n\n")
+}

leetcode/1091.Shortest-Path-in-Binary-Matrix/README.md

@@ -0,0 +1,114 @@
+# [1091. Shortest Path in Binary Matrix](https://leetcode.com/problems/shortest-path-in-binary-matrix/)
+
+
+## 题目
+
+In an N by N square grid, each cell is either empty (0) or blocked (1).
+
+A *clear path from top-left to bottom-right* has length `k` if and only if it is composed of cells `C_1, C_2, ..., C_k` such that:
+
+- Adjacent cells `C_i` and `C_{i+1}` are connected 8-directionally (ie., they are different and share an edge or corner)
+- `C_1` is at location `(0, 0)` (ie. has value `grid[0][0]`)
+- `C_k` is at location `(N-1, N-1)` (ie. has value `grid[N-1][N-1]`)
+- If `C_i` is located at `(r, c)`, then `grid[r][c]` is empty (ie. `grid[r][c] == 0`).
+
+Return the length of the shortest such clear path from top-left to bottom-right.  If such a path does not exist, return -1.
+
+**Example 1:**
+
+```
+Input: [[0,1],[1,0]]
+Output: 2
+```
+
+![https://assets.leetcode.com/uploads/2019/08/04/example1_1.png](https://assets.leetcode.com/uploads/2019/08/04/example1_1.png)
+
+![https://assets.leetcode.com/uploads/2019/08/04/example1_2.png](https://assets.leetcode.com/uploads/2019/08/04/example1_2.png)
+
+**Example 2:**
+
+```
+Input: [[0,0,0],[1,1,0],[1,1,0]]
+Output: 4
+```
+
+![https://assets.leetcode.com/uploads/2019/08/04/example2_1.png](https://assets.leetcode.com/uploads/2019/08/04/example2_1.png)
+
+![https://assets.leetcode.com/uploads/2019/08/04/example2_2.png](https://assets.leetcode.com/uploads/2019/08/04/example2_2.png)
+
+**Note:**
+
+1. `1 <= grid.length == grid[0].length <= 100`
+2. `grid[r][c]` is `0` or `1`
+
+## 题目大意
+
+在一个 N × N 的方形网格中，每个单元格有两种状态：空（0）或者阻塞（1）。一条从左上角到右下角、长度为 k 的畅通路径，由满足下述条件的单元格 C_1, C_2, ..., C_k 组成：
+
+- 相邻单元格 C_i 和 C_{i+1} 在八个方向之一上连通（此时，C_i 和 C_{i+1} 不同且共享边或角）
+- C_1 位于 (0, 0)（即，值为 grid[0][0]）
+- C_k 位于 (N-1, N-1)（即，值为 grid[N-1][N-1]）
+- 如果 C_i 位于 (r, c)，则 grid[r][c] 为空（即，grid[r][c] == 0）
+
+返回这条从左上角到右下角的最短畅通路径的长度。如果不存在这样的路径，返回 -1 。
+
+## 解题思路
+
+- 这一题是简单的找最短路径。利用 BFS 从左上角逐步扩展到右下角，便可以很容易求解。注意每轮扩展需要考虑 8 个方向。
+
+## 代码
+
+```go
+var dir = [][]int{
+	{-1, -1},
+	{-1, 0},
+	{-1, 1},
+	{0, 1},
+	{0, -1},
+	{1, -1},
+	{1, 0},
+	{1, 1},
+}
+
+func shortestPathBinaryMatrix(grid [][]int) int {
+	visited := make([][]bool, 0)
+	for range make([]int, len(grid)) {
+		visited = append(visited, make([]bool, len(grid[0])))
+	}
+	dis := make([][]int, 0)
+	for range make([]int, len(grid)) {
+		dis = append(dis, make([]int, len(grid[0])))
+	}
+	if grid[0][0] == 1 {
+		return -1
+	}
+	if len(grid) == 1 && len(grid[0]) == 1 {
+		return 1
+	}
+
+	queue := []int{0}
+	visited[0][0], dis[0][0] = true, 1
+	for len(queue) > 0 {
+		cur := queue[0]
+		queue = queue[1:]
+		curx, cury := cur/len(grid[0]), cur%len(grid[0])
+		for d := 0; d < 8; d++ {
+			nextx := curx + dir[d][0]
+			nexty := cury + dir[d][1]
+			if isInBoard(grid, nextx, nexty) && !visited[nextx][nexty] && grid[nextx][nexty] == 0 {
+				queue = append(queue, nextx*len(grid[0])+nexty)
+				visited[nextx][nexty] = true
+				dis[nextx][nexty] = dis[curx][cury] + 1
+				if nextx == len(grid)-1 && nexty == len(grid[0])-1 {
+					return dis[nextx][nexty]
+				}
+			}
+		}
+	}
+	return -1
+}
+
+func isInBoard(board [][]int, x, y int) bool {
+	return x >= 0 && x < len(board) && y >= 0 && y < len(board[0])
+}
+```