算法训练 Day 30

LeetCode 51. N 皇后

题目链接：51. N 皇后

思路：回溯经典题，是那种没吃过猪肉也肯定听说过的知名题。其实这道题思路不复杂（相比其他经典的hard题例如接水雨来说），作为困难题主要难在逻辑判断的复杂性、对复杂情况和用例的处理上，此题主要难在皇后是否被攻击的逻辑判断上。还有一些变量的小细节在下方代码中有所注释。

Python版本：

class Solution:
    def solveNQueens(self, n: int) -> List[List[str]]:
        res = []
        chess = [["."]*n for _ in range(n)]

        def backtrack(row):
            if row==n:
                tmp = []
                for path in chess:
                    cur_path = "".join(path)
                    tmp.append(cur_path)
                res.append(tmp)

            # 遍历所在行的每一列
            for i in range(n):
                if self.isValid(n, row, i, chess):
                    chess[row][i] = "Q"
                    backtrack(row+1)  # 此处递归的是下一行, 所以是row+1, 而不是i+1
                    chess[row][i] = "."
        
        backtrack(0)
        return res

    def isValid(self, n, row, col, chess):
        # 判断当前列之上是否已经有皇后
        for i in range(row):
            if chess[i][col] == "Q":
                return False

        # 判断左斜上方是否已经有皇后
        i = row-1
        j = col-1
        while i>=0 and j>=0:
            if chess[i][j] == "Q":
                return False
            i -= 1
            j -= 1

        # 判断右斜上方是否已经有皇后
        i = row-1
        j = col+1
        while i>=0 and j<n:
            if chess[i][j] == "Q":
                return False
            i -= 1
            j += 1

        return True    

go版本：

func solveNQueens(n int) [][]string {
    var res [][]string
    chess := make([][]string, n)
    for i:= 0; i < n; i++ {
		chess[i] = make([]string, n)
	}
	for i := 0; i < n; i++ {
		for j := 0; j < n; j++ {
			chess[i][j] = "."
		}
	}

    var backtrack func(int)
    backtrack = func(row int) {
        if row==n {
            tmp := make([]string, n)
            for i, v := range chess {
				tmp[i] = strings.Join(v, "")
			}
			res = append(res, tmp)
        }
        for i:=0; i<n; i++ {
            if isValid(n, row, i, chess) {
                chess[row][i] = "Q"
				backtrack(row + 1)
				chess[row][i] = "."
            }
        }
    }

    backtrack(0)
	return res
}

func isValid(n, row, col int, chess [][]string) bool {
	for i := 0; i < row; i++ {
		if chess[i][col] == "Q" {
			return false
		}
	}
	for i, j := row-1, col-1; i >= 0 && j >= 0; i, j = i-1, j-1 {
		if chess[i][j] == "Q" {
			return false
		}
	}
	for i, j := row-1, col+1; i >= 0 && j < n; i, j = i-1, j+1 {
		if chess[i][j] == "Q" {
			return false
		}
	}
	return true
}