leetcode刷题

算法/leetcode/Q807 保持城市天际线(Max Increase to Keep City Skyline).md

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+---
+id="2019-01-02-17-15"
+title="Q807 保持城市天际线(Max Increase to Keep City Skyline)"
+headWord="leetcode刷题，题目详情见<a href='https://leetcode-cn.com/problems/max-increase-to-keep-city-skyline/'>点击跳转</a>"
+tags=["java", "leetcode","array",'leetcode']
+category="算法"
+serie="leetcode刷题"
+---
+
+### 解析思路
+
+&emsp;&emsp;leetcode 中等难度中比较简单的一个，题目描述[看这里](https://leetcode-cn.com/problems/max-increase-to-keep-city-skyline/),最开始看题目是比较懵逼的，第一遍没有看懂，多看几次才明白，其实就是让每个点增大到不超过本行且本列的最大值，输出增加的值的和。so 解题方法就是算出每行每列的最大值，然后每个点与当前行当前列最大值中较小的一个差值的和就是结果。
+
+### 代码(Java实现)
+
+```java
+class Solution {
+    public int maxIncreaseKeepingSkyline(int[][] grid) {
+        int rowLength = grid.length;
+        int columnLength = grid[0].length;
+        //每行对应的最大值
+        int[] rowMax = new int[rowLength];
+        //每列对应的最大值
+        Arrays.fill(rowMax, -1);
+        int[] columnMax = new int[columnLength];
+        Arrays.fill(columnMax, -1);
+        int sum = 0;
+        for (int i = 0; i < rowLength; i++) {
+            for (int j = 0; j < columnLength; j++) {
+                if (rowMax[i] == -1) {
+                    //计算行最大值
+                    for (int temp = 0; temp < columnLength; temp++) {
+                        if (rowMax[i] < grid[i][temp]) {
+                            rowMax[i] = grid[i][temp];
+                        }
+                    }
+                }
+                if (columnMax[j] == -1) {
+                    //计算列最大值
+                    for (int temp = 0; temp < rowLength; temp++) {
+                        if (columnMax[j] < grid[temp][j]) {
+                            columnMax[j] = grid[temp][j];
+                        }
+                    }
+                }
+                sum += Math.min(rowMax[i], columnMax[j]) - grid[i][j];
+            }
+        }
+        return sum;
+    }
+}
+```