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| 1 | +#### 1605. 给定行和列的和求可行矩阵 |
| 2 | + |
| 3 | +难度:中等 |
| 4 | + |
| 5 | +--- |
| 6 | + |
| 7 | +给你两个非负整数数组 `rowSum` 和 `colSum` ,其中 `rowSum[i]` 是二维矩阵中第 `i` 行元素的和, `colSum[j]` 是第 `j` 列元素的和。换言之你不知道矩阵里的每个元素,但是你知道每一行和每一列的和。 |
| 8 | + |
| 9 | +请找到大小为 `rowSum.length x colSum.length` 的任意 **非负整数** 矩阵,且该矩阵满足 `rowSum` 和 `colSum` 的要求。 |
| 10 | + |
| 11 | +请你返回任意一个满足题目要求的二维矩阵,题目保证存在 **至少一个** 可行矩阵。 |
| 12 | + |
| 13 | + **示例 1:** |
| 14 | + |
| 15 | +``` |
| 16 | +输入:rowSum = [3,8], colSum = [4,7] |
| 17 | +输出:[[3,0], |
| 18 | + [1,7]] |
| 19 | +解释: |
| 20 | +第 0 行:3 + 0 = 3 == rowSum[0] |
| 21 | +第 1 行:1 + 7 = 8 == rowSum[1] |
| 22 | +第 0 列:3 + 1 = 4 == colSum[0] |
| 23 | +第 1 列:0 + 7 = 7 == colSum[1] |
| 24 | +行和列的和都满足题目要求,且所有矩阵元素都是非负的。 |
| 25 | +另一个可行的矩阵为:[[1,2], |
| 26 | + [3,5]] |
| 27 | +``` |
| 28 | + |
| 29 | + **示例 2:** |
| 30 | + |
| 31 | +``` |
| 32 | +输入:rowSum = [5,7,10], colSum = [8,6,8] |
| 33 | +输出:[[0,5,0], |
| 34 | + [6,1,0], |
| 35 | + [2,0,8]] |
| 36 | +``` |
| 37 | + |
| 38 | + **示例 3:** |
| 39 | + |
| 40 | +``` |
| 41 | +输入:rowSum = [14,9], colSum = [6,9,8] |
| 42 | +输出:[[0,9,5], |
| 43 | + [6,0,3]] |
| 44 | +``` |
| 45 | + |
| 46 | + **示例 4:** |
| 47 | + |
| 48 | +``` |
| 49 | +输入:rowSum = [1,0], colSum = [1] |
| 50 | +输出:[[1], |
| 51 | + [0]] |
| 52 | +``` |
| 53 | + |
| 54 | + **示例 5:** |
| 55 | + |
| 56 | +``` |
| 57 | +输入:rowSum = [0], colSum = [0] |
| 58 | +输出:[[0]] |
| 59 | +``` |
| 60 | + |
| 61 | + **提示:** |
| 62 | + |
| 63 | +* `1 <= rowSum.length, colSum.length <= 500` |
| 64 | +* `0 <= rowSum[i], colSum[i] <= 10^8` |
| 65 | +* `sum(rowSum) == sum(colSum)` |
| 66 | + |
| 67 | +--- |
| 68 | + |
| 69 | +贪心: |
| 70 | + |
| 71 | +运筹学中的运输问题。答案不唯一。有西北角法、最小元素法等,以下是西北角法的代码,即从**左上角**开始分配。 |
| 72 | + |
| 73 | +```Java |
| 74 | +class Solution { |
| 75 | + public int[][] restoreMatrix(int[] rowSum, int[] colSum) { |
| 76 | + int i = 0, j = 0; |
| 77 | + int[][] ans = new int[rowSum.length][colSum.length]; |
| 78 | + while(i < rowSum.length && j < colSum.length){ |
| 79 | + int temp = Math.min(rowSum[i], colSum[j]); |
| 80 | + ans[i][j] = temp; |
| 81 | + rowSum[i] -= temp; |
| 82 | + colSum[j] -= temp; |
| 83 | + if(rowSum[i] == 0) i++; |
| 84 | + if(colSum[j] == 0) j++; |
| 85 | + } |
| 86 | + return ans; |
| 87 | + } |
| 88 | +} |
| 89 | +``` |
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