題意：
尋找最大島。
leetcode.com/problems/ma…
傳入：
[[0,0,1,0,0,0,0,1,0,0,0,0,0],
?[0,0,0,0,0,0,0,1,1,1,0,0,0],
?[0,1,1,0,1,0,0,0,0,0,0,0,0],
?[0,1,0,0,1,1,0,0,1,0,1,0,0],
?[0,1,0,0,1,1,0,0,1,1,1,0,0],
?[0,0,0,0,0,0,0,0,0,0,1,0,0],
?[0,0,0,0,0,0,0,1,1,1,0,0,0],
?[0,0,0,0,0,0,0,1,1,0,0,0,0]]

傳出：
6

思路：
從左上角開始一個個往右下角數(shù)，每一個數(shù)出來連塊最大數(shù)，然后輸出。
連塊最大數(shù)計算：按照 上、右、下、左 方向，取得連著的值，推入暫存數(shù)組，再對連著的值進行遞歸調(diào)用，再推入暫存數(shù)組。

代碼：

/** * @param {number[][]} grid * @return {number} */ var maxAreaOfIsland = function(grid) {function add(list, i , j){if(!list)return;for(var k = 0; k < list.length; ++k){if(list[k].i === i && list[k].j === j)return false;}list.push({i, j});return true;}function deepSearch(i, j, list){if(!list)list = [];if(grid[i][j]){add(list, i, j);}if(j > 0 && grid[i][j - 1]){if(add(list, i, j - 1)){deepSearch(i, j - 1, list);}}if(i < grid.length - 1 && grid[i + 1][j]){if(add(list, i + 1, j)){deepSearch(i + 1, j, list);}}if(grid.length && j < grid[0].length - 1 && grid[i][j + 1]){if(add(list, i, j + 1)){deepSearch(i, j + 1, list);}}if(i > 0 && grid[i - 1][j]){if(add(list, i - 1, j)){deepSearch(i - 1, j, list);}}return list.length;}function countArea(i, j){if(j > 0 && grid[i][j - 1]) // 不是最左上角的不countreturn 0;if(i > 0 && grid[i - 1][j])return 0;return deepSearch(i, j);}var max = 0;for(var i = 0;i < grid.length; ++i){for(var j = 0;j < grid[i].length; ++j){var obj = grid[i][j];if(obj){var tmp = countArea(i, j);max = tmp > max ? tmp : max;}}}return max; }; 復(fù)制代碼

轉(zhuǎn)載于:https://juejin.im/post/5a32b03f5188250648693d76

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