update: var name & make

This commit is contained in:
huoyinghui
2021-12-17 11:27:48 +08:00
committed by tphyhFighting
parent ce7c41b9c4
commit e3bde6844b
2 changed files with 12 additions and 10 deletions

View File

@ -1,8 +1,9 @@
package leetcode
func maxIncreaseKeepingSkyline(grid [][]int) int {
var topBottomSkyline []int
var leftRightSKyline []int
n := len(grid)
topBottomSkyline := make([]int, 0, n)
leftRightSkyline := make([]int, 0, n)
for i := range grid {
cur := 0
for _, v := range grid[i] {
@ -10,7 +11,7 @@ func maxIncreaseKeepingSkyline(grid [][]int) int {
cur = v
}
}
leftRightSKyline = append(leftRightSKyline, cur)
leftRightSkyline = append(leftRightSkyline, cur)
}
for j := range grid {
cur := 0
@ -24,7 +25,7 @@ func maxIncreaseKeepingSkyline(grid [][]int) int {
var ans int
for i := range grid {
for j := 0; j < len(grid[0]); j++ {
ans += min(topBottomSkyline[j], leftRightSKyline[i]) - grid[i][j]
ans += min(topBottomSkyline[j], leftRightSkyline[i]) - grid[i][j]
}
}
return ans

View File

@ -48,8 +48,8 @@ Return the maximum total sum that the height of the buildings can be increased b
## 解题思路
- 从数组竖直方向即顶部底部看“天际线”计算出topBottomSkyline
- 从数组水平方向即左侧右侧看“天际线”计算出leftRightSKyline
- 计算grid中每个元素与对应的topBottomSkyline和leftRightSKyline中较小值的差值
- 从数组水平方向即左侧右侧看“天际线”计算出leftRightSkyline
- 计算grid中每个元素与对应的topBottomSkyline和leftRightSkyline中较小值的差值
- 统计所有差值的总和ans并返回
## 代码
@ -58,8 +58,9 @@ Return the maximum total sum that the height of the buildings can be increased b
package leetcode
func maxIncreaseKeepingSkyline(grid [][]int) int {
var topBottomSkyline []int
var leftRightSKyline []int
n := len(grid)
topBottomSkyline := make([]int, 0, n)
leftRightSkyline := make([]int, 0, n)
for i := range grid {
cur := 0
for _, v := range grid[i] {
@ -67,7 +68,7 @@ func maxIncreaseKeepingSkyline(grid [][]int) int {
cur = v
}
}
leftRightSKyline = append(leftRightSKyline, cur)
leftRightSkyline = append(leftRightSkyline, cur)
}
for j := range grid {
cur := 0
@ -81,7 +82,7 @@ func maxIncreaseKeepingSkyline(grid [][]int) int {
var ans int
for i := range grid {
for j := 0; j < len(grid[0]); j++ {
ans += min(topBottomSkyline[j], leftRightSKyline[i]) - grid[i][j]
ans += min(topBottomSkyline[j], leftRightSkyline[i]) - grid[i][j]
}
}
return ans