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64. 最小路径和 #51

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@Geekhyt

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@Geekhyt

原题链接

状态定义

dp[i][j] 代表从 (0, 0) 走到 (i, j) 的最小路径和。

状态转移方程

明确题意:只能向右或者向下走,也就是说终点 (i, j) 只能从 (i - 1, j) 或者 (i, j - 1) 走过来。

分为以下三种情况分别处理:

  1. 左边或者上边都是边界时,终点也就是起点。
  • dp[i][j] = grid[i][j]
  1. 左边或者上边为边界时:
  • grid[i][j] = grid[i - 1][j] + grid[i][j]
  • grid[i][j] = grid[i][j - 1] + grid[i][j]
  1. 左边和上边都不为边界时:
  • grid[i][j] = Math.min(grid[i - 1][j], grid[i][j - 1]) + grid[i][j]
constminPathSum=function(grid){constn=grid.lengthconstm=grid[0].lengthconstdp=Array.from(newArray(n),()=>newArray(m))for(leti=0;i<n;i++){for(letj=0;j<m;j++){if(i!==0&&j!==0){dp[i][j]=Math.min(dp[i-1][j],dp[i][j-1])+grid[i][j]}elseif(i!==0&&j===0){dp[i][j]=dp[i-1][j]+grid[i][j]}elseif(i===0&&j!==0){dp[i][j]=dp[i][j-1]+grid[i][j]}elseif(i===0&&j===0){dp[i][j]=grid[i][j]}}}returndp[n-1][m-1]}
  • 时间复杂度: O(m * n)
  • 空间复杂度: O(m * n)

降维

不需要额外建立空间,直接遍历修改grid[i][j]

constminPathSum=function(grid){constn=grid.lengthconstm=grid[0].lengthfor(leti=0;i<n;i++){for(letj=0;j<m;j++){if(i!==0&&j!==0){grid[i][j]=Math.min(grid[i-1][j],grid[i][j-1])+grid[i][j]}elseif(i!==0&&j===0){grid[i][j]=grid[i-1][j]+grid[i][j]}elseif(i===0&&j!==0){grid[i][j]=grid[i][j-1]+grid[i][j]}elseif(i===0&&j===0){grid[i][j]=grid[i][j]}}}returngrid[n-1][m-1]}
  • 时间复杂度: O(m * n)
  • 空间复杂度: O(1)

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