Open Shawngbk opened 7 years ago
思路：第二while循环找出了距离第i个房子最近的heater，第二个循环找的是最大的半径，这样才能选的heater数量最少 public class Solution { public int findRadius(int[] houses, int[] heaters) { Arrays.sort(houses); Arrays.sort(heaters); int i = 0, j = 0, res = 0;//i is pointer of houses, j is pointer pf heaters while(i < houses.length) { while(j < heaters.length - 1 && Math.abs(heaters[j+1]-houses[i]) <= Math.abs(heaters[j]-houses[i])) { j++; } res = Math.max(res, Math.abs(heaters[j]-houses[i])); i++; } return res; } }
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Shawngbk
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7 years ago Shawngbk
commented
7 years ago
思路：第二while循环找出了距离第i个房子最近的heater，第二个循环找的是最大的半径，这样才能选的heater数量最少 public class Solution { public int findRadius(int[] houses, int[] heaters) { Arrays.sort(houses); Arrays.sort(heaters); int i = 0, j = 0, res = 0;//i is pointer of houses, j is pointer pf heaters while(i < houses.length) { while(j < heaters.length - 1 && Math.abs(heaters[j+1]-houses[i]) <= Math.abs(heaters[j]-houses[i])) { j++; } res = Math.max(res, Math.abs(heaters[j]-houses[i])); i++; } return res; } }