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https://github.com/halfrost/LeetCode-Go.git
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update: var name & make
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committed by
tphyhFighting

parent
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commit
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@ -1,8 +1,9 @@
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package leetcode
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func maxIncreaseKeepingSkyline(grid [][]int) int {
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var topBottomSkyline []int
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var leftRightSKyline []int
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n := len(grid)
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topBottomSkyline := make([]int, 0, n)
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leftRightSkyline := make([]int, 0, n)
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for i := range grid {
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cur := 0
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for _, v := range grid[i] {
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@ -10,7 +11,7 @@ func maxIncreaseKeepingSkyline(grid [][]int) int {
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cur = v
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}
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}
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leftRightSKyline = append(leftRightSKyline, cur)
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leftRightSkyline = append(leftRightSkyline, cur)
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}
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for j := range grid {
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cur := 0
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@ -24,7 +25,7 @@ func maxIncreaseKeepingSkyline(grid [][]int) int {
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var ans int
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for i := range grid {
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for j := 0; j < len(grid[0]); j++ {
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ans += min(topBottomSkyline[j], leftRightSKyline[i]) - grid[i][j]
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ans += min(topBottomSkyline[j], leftRightSkyline[i]) - grid[i][j]
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}
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}
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return ans
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@ -48,8 +48,8 @@ Return the maximum total sum that the height of the buildings can be increased b
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## 解题思路
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- 从数组竖直方向(即顶部,底部)看“天际线”计算出topBottomSkyline
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- 从数组水平方向(即左侧,右侧)看“天际线”计算出leftRightSKyline
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- 计算grid中每个元素与对应的topBottomSkyline和leftRightSKyline中较小值的差值
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- 从数组水平方向(即左侧,右侧)看“天际线”计算出leftRightSkyline
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- 计算grid中每个元素与对应的topBottomSkyline和leftRightSkyline中较小值的差值
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- 统计所有差值的总和ans并返回
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## 代码
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@ -58,8 +58,9 @@ Return the maximum total sum that the height of the buildings can be increased b
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package leetcode
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func maxIncreaseKeepingSkyline(grid [][]int) int {
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var topBottomSkyline []int
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var leftRightSKyline []int
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n := len(grid)
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topBottomSkyline := make([]int, 0, n)
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leftRightSkyline := make([]int, 0, n)
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for i := range grid {
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cur := 0
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for _, v := range grid[i] {
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@ -67,7 +68,7 @@ func maxIncreaseKeepingSkyline(grid [][]int) int {
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cur = v
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}
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}
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leftRightSKyline = append(leftRightSKyline, cur)
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leftRightSkyline = append(leftRightSkyline, cur)
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}
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for j := range grid {
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cur := 0
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@ -81,7 +82,7 @@ func maxIncreaseKeepingSkyline(grid [][]int) int {
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var ans int
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for i := range grid {
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for j := 0; j < len(grid[0]); j++ {
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ans += min(topBottomSkyline[j], leftRightSKyline[i]) - grid[i][j]
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ans += min(topBottomSkyline[j], leftRightSkyline[i]) - grid[i][j]
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}
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}
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return ans
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