leetcode刷题记录 5



Triangle

思路:

state: f[i][j]表示从i,j出发到最后一层的最小路径和

function: f[i][j] = min(f[i+1][j], f[i+1][j+1]) + triangle[i][j]

initialize: f[n-1][j] = triangle[n-1][j]

answer: f[0][0]

可省去f[i][j]第一维空间

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public class  {
public int minimumTotal(List<List<Integer>> triangle) {
int n = triangle.size();
int[] sum = new int[n];


for (int i = 0; i < n; i++) {
sum[i] = triangle.get(n - 1).get(i);
}

// 自底向上递推,更新sum[]
for (int i = n - 2; i >= 0; i--) {
for (int j = 0; j <= i; j++) {
sum[j] = triangle.get(i).get(j) + Math.min(sum[j], sum[j + 1]);
}
}

return sum[0];
}
}


Unique Paths

思路:

state: f[i][j]表示从起点到i,j的路径数

function: f[i][j] = f[i-1][j] + f[i][j-1]

initialize: f[0][i] = 1, f[j][0] = 1

answer: f[m-1][n-1]

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public class  {
public int uniquePaths(int m, int n) {
int[][] grid = new int[m][n];
for (int i = 0; i < n; i++) {
grid[0][i] = 1;
}
for (int i = 0; i < m; i++) {
grid[i][0] = 1;
}
for (int i = 1; i < m; i++) {
for (int j = 1; j < n; j++) {
grid[i][j] = grid[i - 1][j] + grid[i][j - 1];
}
}
return grid[m - 1][n - 1];
}
}


Unique Paths II

思路:

state: f[i][j]表示从起点到i,j的路径数

function: f[i][j] = f[i-1][j] + f[i][j-1]

initialize: f[0][i] = 1, f[j][0] = 1

answer: f[m-1][n-1]

obstacleGrid[i][j]为1时表示路径不通,则grid[i][j] = 0

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public class  {
public int uniquePathsWithObstacles(int[][] obstacleGrid) {
int m = obstacleGrid.length;
int n = obstacleGrid[0].length;
int[][] grid = new int[m][n];

for (int i = 0; i < m; i++) {
if (obstacleGrid[i][0] == 1) {
break;
}
grid[i][0] = 1;
}
for (int j = 0; j < n; j++) {
if (obstacleGrid[0][j] == 1) {
break;
}
grid[0][j] = 1;
}

for (int i = 1; i < m; i++) {
for (int j = 1; j < n; j++) {
if (obstacleGrid[i][j] == 1) {
grid[i][j] = 0;
} else {
grid[i][j] = grid[i - 1][j] + grid[i][j - 1];
}
}
}

return grid[m - 1][n - 1];
}
}