本文主要跟大家分享一個關於javascript二維陣列的面試題,希望能幫助大家。給定一個二維數組,實現一個功能函數fn,向這個函數中傳遞這個二維數組的一個坐標,如果這個坐標的值為”1“,將返回和這個坐標所有相連的並且坐標值為1坐標。
例如,傳遞了fn([3,4])所得到的結果為:
[[3,4],[4,4],[5,4],[6,4] ,[7,4],[8,4],[8,5],[8,6]]
var arr =[
[0,0,0,0,0,0,0,0,0,0,0],
[0,0,0,0,0,0,0,0,0,0,0],
[0,0,0,0,0,0,0,0,0,0,0],
[0,0,0,0,1,0,0,0,1,0,0],
[0,0,0,0,1,0,0,0,1,0,0],
[0,0,0,0,1,0,0,0,1,0,0],
[0,0,0,0,1,0,0,0,0,0,0],
[0,0,0,0,1,0,0,0,0,0,0],
[0,0,0,0,1,1,1,0,0,0,0],
[0,0,0,0,0,0,0,0,0,0,0],
[0,0,0,0,0,0,0,0,0,0,0],
] ;
解答想法:寬度優先遍歷即可。供參考,相連條件沒給出且當做是橫豎方向:
var arr =[
[0,0,0,0,0,0,0,0,0,0,0],
[0,0,0,0,0,0,0,0,0,0,0],
[0,0,0,0,0,0,0,0,0,0,0],
[0,0,0,0,1,0,0,0,1,0,0],
[0,0,0,0,1,0,0,0,1,0,0],
[0,0,0,0,1,0,0,0,1,0,0],
[0,0,0,0,1,0,0,0,0,0,0],
[0,0,0,0,1,0,0,0,0,0,0],
[0,0,0,0,1,1,1,0,0,0,0],
[0,0,0,0,0,0,0,0,0,0,0],
[0,0,0,0,0,0,0,0,0,0,0],
]
function fn ([x, y]) {
if (arr[x][y] !== 1) return false
const queue = [[x, y]]
const memo = arr.map(row => new Array(row.length).fill(false))
const direction = [
[-1, 0],
[1, 0],
[0, -1],
[0, 1],
]
while(queue.length > 0) {
const [x, y] = queue.pop()
direction.forEach(([h, v]) => {
const newX = x + h
const newY = y + v
if (arr[newX][newY] === 1 && !memo[newX][newY]) {
memo[newX][newY] = true
queue.push([newX, newY])
}
})
}
const result = []
for (let i = 0; i < arr.length; i++) {
for (let j = 0; j < arr[i].length; j++) {
if(memo[i][j]) {
result.push([i, j])
}
}
}
return result
}
console.log(fn([3,4]))
出隊列進行新一輪的匹配所以只要再用一個緩存隊列存儲過往匹配的成功數據即可,沒有必要最後在進行遍歷的必要。程式碼如下:var arr =[
function fn(point) {
var memo = {}, result = [], direction = [[-1, 0],[1, 0],[0, -1],[0, 1]]
function dg([x, y]) {
result.push(memo[x + "," + y] = [x, y]);
direction.forEach(([h, v]) => {
const newX = x + h
const newY = y + v
if (arr[newX][newY] === 1 && !memo[newX + "," + newY]) {
dg([newX, newY]);
}
})
}
dg(point);
return result;
}
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