2021-12-28：给定一个二维数组matrix，matrixi = k代表:
从(i,j)位置可以随意往右跳<=k步，或者从(i,j)位置可以随意往下跳<=k步，
如果matrixi = 0，代表来到(i,j)位置必须停止，
返回从matrix左上角到右下角，至少要跳几次，
已知matrix中行数n <= 5000, 列数m <= 5000，
matrix中的值，<= 5000。
来自京东。
答案2021-12-28：
方法一：自然智慧。递归。复杂度过不了。
方法二：动态规划 线段树。
代码用golang编写。代码如下：
代码语言：txt复制package main
import (
"fmt"
"math"
)
func main() {
ret := jump2([][]int{{1, 2, 3, 4, 5}, {6, 7, 8, 9, 10}})
fmt.Println(ret)
}
func getMin(a, b int) int {
if a < b {
return a
} else {
return b
}
}
// 优化方法, 利用线段树做枚举优化
// 因为线段树，下标从1开始
// 所以，该方法中所有的下标，请都从1开始，防止乱！
func jump2(arr [][]int) int {
n := len(arr)
m := len(arr[0])
map0 := make([][]int, n 1)
for i := 0; i < n 1; i {
map0[i] = make([]int, m 1)
}
for a, b := 0, 1; a < n; a, b = a 1, b 1 {
for c, d := 0, 1; c < m; c, d = c 1, d 1 {
map0[b][d] = arr[a][c]
}
}
rowTrees := make([]*SegmentTree, n 1)
for i := 1; i <= n; i {
rowTrees[i] = NewSegmentTree(m)
}
colTrees := make([]*SegmentTree, m 1)
for i := 1; i <= m; i {
colTrees[i] = NewSegmentTree(n)
}
rowTrees[n].update0(m, m, 0, 1, m, 1)
colTrees[m].update0(n, n, 0, 1, n, 1)
for col := m - 1; col >= 1; col-- {
if map0[n][col] != 0 {
left := col 1
right := getMin(col map0[n][col], m)
next := rowTrees[n].query(left, right, 1, m, 1)
if next != math.MaxInt64 {
rowTrees[n].update0(col, col, next 1, 1, m, 1)
colTrees[col].update0(n, n, next 1, 1, n, 1)
}
}
}
for row := n - 1; row >= 1; row-- {
if map0[row][m] != 0 {
up := row 1
down := getMin(row map0[row][m], n)
next := colTrees[m].query(up, down, 1, n, 1)
if next != math.MaxInt64 {
rowTrees[row].update0(m, m, next 1, 1, m, 1)
colTrees[m].update0(row, row, next 1, 1, n, 1)
}
}
}
for row := n - 1; row >= 1; row-- {
for col := m - 1; col >= 1; col-- {
if map0[row][col] != 0 {
// (row,col) 往右是什么范围呢？[left,right]
left := col 1
right := getMin(col map0[row][col], m)
next1 := rowTrees[row].query(left, right, 1, m, 1)
// (row,col) 往下是什么范围呢？[up,down]
up := row 1
down := getMin(row map0[row][col], n)
next2 := colTrees[col].query(up, down, 1, n, 1)
next := getMin(next1, next2)
if next != math.MaxInt64 {
rowTrees[row].update0(col, col, next 1, 1, m, 1)
colTrees[col].update0(row, row, next 1, 1, n, 1)
}
}
}
}
return rowTrees[1].query(1, 1, 1, m, 1)
}
// 区间查询最小值的线段树
// 注意下标从1开始，不从0开始
// 比如你传入size = 8
// 则位置对应为1~8，而不是0~7
type SegmentTree struct {
min []int
change []int
update []bool
}
func NewSegmentTree(size int) *SegmentTree {
ret := &SegmentTree{}
N := size 1
ret.min = make([]int, N<<2)
ret.change = make([]int, N<<2)
ret.update = make([]bool, N<<2)
ret.update0(1, size, math.MaxInt64, 1, size, 1)
return ret
}
func (this *SegmentTree) pushUp(rt int) {
this.min[rt] = getMin(this.min[rt<<1], this.min[rt<<1|1])
}
func (this *SegmentTree) pushDown(rt, ln, rn int) {
if this.update[rt] {
this.update[rt<<1] = true
this.update[rt<<1|1] = true
this.change[rt<<1] = this.change[rt]
this.change[rt<<1|1] = this.change[rt]
this.min[rt<<1] = this.change[rt]
this.min[rt<<1|1] = this.change[rt]
this.update[rt] = false
}
}
// 最后三个参数是固定的, 每次传入相同的值即可:
// l = 1(固定)
// r = size(你设置的线段树大小)
// rt = 1(固定)
func (this *SegmentTree) update0(L, R, C, l, r, rt int) {
if L <= l && r <= R {
this.update[rt] = true
this.change[rt] = C
this.min[rt] = C
return
}
mid := (l r) >> 1
this.pushDown(rt, mid-l 1, r-mid)
if L <= mid {
this.update0(L, R, C, l, mid, rt<<1)
}
if R > mid {
this.update0(L, R, C, mid 1, r, rt<<1|1)
}
this.pushUp(rt)
}
// 最后三个参数是固定的, 每次传入相同的值即可:
// l = 1(固定)
// r = size(你设置的线段树大小)
// rt = 1(固定)
func (this *SegmentTree) query(L, R, l, r, rt int) int {
if L <= l && r <= R {
return this.min[rt]
}
mid := (l r) >> 1
this.pushDown(rt, mid-l 1, r-mid)
left := math.MaxInt64
right := math.MaxInt64
if L <= mid {
left = this.query(L, R, l, mid, rt<<1)
}
if R > mid {
right = this.query(L, R, mid 1, r, rt<<1|1)
}
return getMin(left, right)
}执行结果如下：
图片左神java代码
